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Similarity of Triangles and Basic Proportionality Theorem Questions in English
Class 10 Mathematics · Triangles · Similarity of Triangles and Basic Proportionality Theorem
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51
Difficult
In the given figure,if $PQRS$ is a parallelogram and $AB \parallel PS$,then prove that $OC \parallel SR$.
Solution
(N/A) Given: $PQRS$ is a parallelogram,so $PQ \parallel SR$ and $PS \parallel QR$. Also,$AB \parallel PS$.
To prove: $OC \parallel SR$.
Proof: In $\triangle OPS$ and $\triangle OAB$,since $PS \parallel AB$:
$\angle POS = \angle AOB$ (Common angle)
$\angle OSP = \angle OBA$ (Corresponding angles)
Therefore,$\triangle OPS \sim \triangle OAB$ (by $AAA$ similarity criterion).
Thus,$\frac{PS}{AB} = \frac{OS}{OB}$ ..........$(i)$
In $\triangle CQR$ and $\triangle CAB$,since $QR \parallel PS \parallel AB$:
$\angle QCR = \angle ACB$ (Common angle)
$\angle CQR = \angle CAB$ (Corresponding angles)
Therefore,$\triangle CQR \sim \triangle CAB$ (by $AA$ similarity criterion).
Thus,$\frac{QR}{AB} = \frac{CR}{CB}$.
Since $PQRS$ is a parallelogram,$PS = QR$. Substituting this,we get:
$\frac{PS}{AB} = \frac{CR}{CB}$ ..........$(ii)$
From equations $(i)$ and $(ii)$:
$\frac{OS}{OB} = \frac{CR}{CB} \Rightarrow \frac{OB}{OS} = \frac{CB}{CR}$.
Subtracting $1$ from both sides:
$\frac{OB}{OS} - 1 = \frac{CB}{CR} - 1$
$\frac{OB - OS}{OS} = \frac{CB - CR}{CR}$
$\frac{BS}{OS} = \frac{BR}{CR}$.
By the converse of the Basic Proportionality Theorem $(BPT)$,$SR \parallel OC$ (or $OC \parallel SR$). Hence proved.
To prove: $OC \parallel SR$.
Proof: In $\triangle OPS$ and $\triangle OAB$,since $PS \parallel AB$:
$\angle POS = \angle AOB$ (Common angle)
$\angle OSP = \angle OBA$ (Corresponding angles)
Therefore,$\triangle OPS \sim \triangle OAB$ (by $AAA$ similarity criterion).
Thus,$\frac{PS}{AB} = \frac{OS}{OB}$ ..........$(i)$
In $\triangle CQR$ and $\triangle CAB$,since $QR \parallel PS \parallel AB$:
$\angle QCR = \angle ACB$ (Common angle)
$\angle CQR = \angle CAB$ (Corresponding angles)
Therefore,$\triangle CQR \sim \triangle CAB$ (by $AA$ similarity criterion).
Thus,$\frac{QR}{AB} = \frac{CR}{CB}$.
Since $PQRS$ is a parallelogram,$PS = QR$. Substituting this,we get:
$\frac{PS}{AB} = \frac{CR}{CB}$ ..........$(ii)$
From equations $(i)$ and $(ii)$:
$\frac{OS}{OB} = \frac{CR}{CB} \Rightarrow \frac{OB}{OS} = \frac{CB}{CR}$.
Subtracting $1$ from both sides:
$\frac{OB}{OS} - 1 = \frac{CB}{CR} - 1$
$\frac{OB - OS}{OS} = \frac{CB - CR}{CR}$
$\frac{BS}{OS} = \frac{BR}{CR}$.
By the converse of the Basic Proportionality Theorem $(BPT)$,$SR \parallel OC$ (or $OC \parallel SR$). Hence proved.
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View Solution52
Difficult
In the figure,$PA$,$QB$,$RC$,and $SD$ are all perpendiculars to a line $l$. Given $AB = 6 \, cm$,$BC = 9 \, cm$,$CD = 12 \, cm$,and $PS = 36 \, cm$. Find $PQ$,$QR$,and $RS$.
Solution
(N/A) Given: $AB = 6 \, cm$,$BC = 9 \, cm$,$CD = 12 \, cm$,and $PS = 36 \, cm$.
Also,$PA$,$QB$,$RC$,and $SD$ are all perpendiculars to line $l$,which implies $PA \parallel QB \parallel RC \parallel SD$.
By the intercept theorem (or Thales' theorem application for parallel lines),the ratio of intercepts on one transversal is equal to the ratio of intercepts on another transversal:
$PQ : QR : RS = AB : BC : CD$
$PQ : QR : RS = 6 : 9 : 12$
Let $PQ = 6x$,$QR = 9x$,and $RS = 12x$.
Since the total length $PS = 36 \, cm$:
$PQ + QR + RS = 36$
$6x + 9x + 12x = 36$
$27x = 36$
$x = \frac{36}{27} = \frac{4}{3}$
Now,calculating the lengths:
$PQ = 6x = 6 \times \frac{4}{3} = 8 \, cm$
$QR = 9x = 9 \times \frac{4}{3} = 12 \, cm$
$RS = 12x = 12 \times \frac{4}{3} = 16 \, cm$
Also,$PA$,$QB$,$RC$,and $SD$ are all perpendiculars to line $l$,which implies $PA \parallel QB \parallel RC \parallel SD$.
By the intercept theorem (or Thales' theorem application for parallel lines),the ratio of intercepts on one transversal is equal to the ratio of intercepts on another transversal:
$PQ : QR : RS = AB : BC : CD$
$PQ : QR : RS = 6 : 9 : 12$
Let $PQ = 6x$,$QR = 9x$,and $RS = 12x$.
Since the total length $PS = 36 \, cm$:
$PQ + QR + RS = 36$
$6x + 9x + 12x = 36$
$27x = 36$
$x = \frac{36}{27} = \frac{4}{3}$
Now,calculating the lengths:
$PQ = 6x = 6 \times \frac{4}{3} = 8 \, cm$
$QR = 9x = 9 \times \frac{4}{3} = 12 \, cm$
$RS = 12x = 12 \times \frac{4}{3} = 16 \, cm$
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View Solution53
Difficult
$O$ is the point of intersection of the diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$. Through $O$,a line segment $PQ$ is drawn parallel to $AB$ meeting $AD$ in $P$ and $BC$ in $Q$. Prove that $PO = QO$.
Solution
(N/A) Given: $ABCD$ is a trapezium with $AB \parallel DC$. Diagonals $AC$ and $BD$ intersect at $O$. $PQ \parallel AB \parallel DC$.
To prove: $PO = QO$.
Proof: In $\triangle ADC$,$PO \parallel DC$ (since $PQ \parallel DC$).
By the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{AP}{PD} = \frac{AO}{OC}$ ........$(i)$
In $\triangle ABC$,$OQ \parallel AB$.
By $BPT$,we have:
$\frac{BQ}{QC} = \frac{AO}{OC}$ ........$(ii)$
From $(i)$ and $(ii)$,we get:
$\frac{AP}{PD} = \frac{BQ}{QC}$
Adding $1$ to both sides:
$\frac{AP}{PD} + 1 = \frac{BQ}{QC} + 1$
$\frac{AP + PD}{PD} = \frac{BQ + QC}{QC}$
$\frac{AD}{PD} = \frac{BC}{QC}$
$\Rightarrow \frac{PD}{AD} = \frac{QC}{BC}$ ........$(iii)$
Now,in $\triangle ABD$,$PO \parallel AB$. Therefore,$\triangle POD \sim \triangle ABD$ (by $AA$ similarity).
So,$\frac{PO}{AB} = \frac{PD}{AD}$ ........$(iv)$
In $\triangle ABC$,$OQ \parallel AB$. Therefore,$\triangle OQC \sim \triangle ABC$ (by $AA$ similarity).
So,$\frac{OQ}{AB} = \frac{QC}{BC}$ ........$(v)$
From $(iii)$,$(iv)$,and $(v)$,we get:
$\frac{PO}{AB} = \frac{OQ}{AB}$
$\Rightarrow PO = OQ$. Hence proved.
To prove: $PO = QO$.
Proof: In $\triangle ADC$,$PO \parallel DC$ (since $PQ \parallel DC$).
By the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{AP}{PD} = \frac{AO}{OC}$ ........$(i)$
In $\triangle ABC$,$OQ \parallel AB$.
By $BPT$,we have:
$\frac{BQ}{QC} = \frac{AO}{OC}$ ........$(ii)$
From $(i)$ and $(ii)$,we get:
$\frac{AP}{PD} = \frac{BQ}{QC}$
Adding $1$ to both sides:
$\frac{AP}{PD} + 1 = \frac{BQ}{QC} + 1$
$\frac{AP + PD}{PD} = \frac{BQ + QC}{QC}$
$\frac{AD}{PD} = \frac{BC}{QC}$
$\Rightarrow \frac{PD}{AD} = \frac{QC}{BC}$ ........$(iii)$
Now,in $\triangle ABD$,$PO \parallel AB$. Therefore,$\triangle POD \sim \triangle ABD$ (by $AA$ similarity).
So,$\frac{PO}{AB} = \frac{PD}{AD}$ ........$(iv)$
In $\triangle ABC$,$OQ \parallel AB$. Therefore,$\triangle OQC \sim \triangle ABC$ (by $AA$ similarity).
So,$\frac{OQ}{AB} = \frac{QC}{BC}$ ........$(v)$
From $(iii)$,$(iv)$,and $(v)$,we get:
$\frac{PO}{AB} = \frac{OQ}{AB}$
$\Rightarrow PO = OQ$. Hence proved.
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View Solution54
Difficult
In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$. If $AP = 3$,$PB = 5$ and $AQ = 6$,find $QC$ and $AC$.
Solution
(N/A) In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$.
By the Basic Proportionality Theorem (Thales Theorem):
$\frac{AP}{PB} = \frac{AQ}{QC}$
Given $AP = 3$,$PB = 5$,$AQ = 6$.
$\frac{3}{5} = \frac{6}{QC}$
$QC = \frac{6 \times 5}{3} = 10$.
Now,to find $AC$:
$AC = AQ + QC = 6 + 10 = 16$.
Alternatively,using the corollary $\frac{AP}{AB} = \frac{AQ}{AC}$:
$\frac{3}{3+5} = \frac{6}{AC}$
$\frac{3}{8} = \frac{6}{AC}$
$AC = \frac{6 \times 8}{3} = 16$.
By the Basic Proportionality Theorem (Thales Theorem):
$\frac{AP}{PB} = \frac{AQ}{QC}$
Given $AP = 3$,$PB = 5$,$AQ = 6$.
$\frac{3}{5} = \frac{6}{QC}$
$QC = \frac{6 \times 5}{3} = 10$.
Now,to find $AC$:
$AC = AQ + QC = 6 + 10 = 16$.
Alternatively,using the corollary $\frac{AP}{AB} = \frac{AQ}{AC}$:
$\frac{3}{3+5} = \frac{6}{AC}$
$\frac{3}{8} = \frac{6}{AC}$
$AC = \frac{6 \times 8}{3} = 16$.
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View Solution55
MediumMCQ
In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $\frac{AB}{5} = \frac{AC}{3}$ and $BD = 4.5$,find $BC$.
A
$4.5$
B
$3.6$
C
$7.2$
D
$6.2$
Solution
(C) In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. According to the Angle Bisector Theorem:
$\frac{AB}{AC} = \frac{BD}{DC}$
Given $\frac{AB}{5} = \frac{AC}{3}$,we have $\frac{AB}{AC} = \frac{5}{3}$.
Substituting the values:
$\frac{5}{3} = \frac{4.5}{DC}$
$DC = \frac{4.5 \times 3}{5} = 0.9 \times 3 = 2.7$
Since $D$ lies on $BC$,$BC = BD + DC$.
$BC = 4.5 + 2.7 = 7.2$.
$\frac{AB}{AC} = \frac{BD}{DC}$
Given $\frac{AB}{5} = \frac{AC}{3}$,we have $\frac{AB}{AC} = \frac{5}{3}$.
Substituting the values:
$\frac{5}{3} = \frac{4.5}{DC}$
$DC = \frac{4.5 \times 3}{5} = 0.9 \times 3 = 2.7$
Since $D$ lies on $BC$,$BC = BD + DC$.
$BC = 4.5 + 2.7 = 7.2$.
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View Solution56
Difficult
In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. Prove that $BD = \frac{BC \times AB}{AB + AC}$ and $DC = \frac{BC \times AC}{AB + AC}$.
Solution
(N/A) In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$.
By the Angle Bisector Theorem,we have:
$\frac{AB}{AC} = \frac{BD}{DC}$
To find $BD$:
$\frac{AB}{AC} = \frac{BD}{DC}$
Using the property of Componendo,$\frac{AB + AC}{AC} = \frac{BD + DC}{DC} = \frac{BC}{DC}$
This does not directly give $BD$. Let's use $\frac{AB}{AC} = \frac{BD}{DC} \implies \frac{AB}{AB + AC} = \frac{BD}{BD + DC} = \frac{BD}{BC}$
$\therefore BD = \frac{BC \times AB}{AB + AC}$
To find $DC$:
$\frac{AB}{AC} = \frac{BD}{DC} \implies \frac{AC}{AB} = \frac{DC}{BD}$
Using Componendo,$\frac{AC + AB}{AB} = \frac{DC + BD}{BD} = \frac{BC}{BD}$
$\implies \frac{AC}{AC + AB} = \frac{DC}{BC}$
$\therefore DC = \frac{BC \times AC}{AB + AC}$
By the Angle Bisector Theorem,we have:
$\frac{AB}{AC} = \frac{BD}{DC}$
To find $BD$:
$\frac{AB}{AC} = \frac{BD}{DC}$
Using the property of Componendo,$\frac{AB + AC}{AC} = \frac{BD + DC}{DC} = \frac{BC}{DC}$
This does not directly give $BD$. Let's use $\frac{AB}{AC} = \frac{BD}{DC} \implies \frac{AB}{AB + AC} = \frac{BD}{BD + DC} = \frac{BD}{BC}$
$\therefore BD = \frac{BC \times AB}{AB + AC}$
To find $DC$:
$\frac{AB}{AC} = \frac{BD}{DC} \implies \frac{AC}{AB} = \frac{DC}{BD}$
Using Componendo,$\frac{AC + AB}{AB} = \frac{DC + BD}{BD} = \frac{BC}{BD}$
$\implies \frac{AC}{AC + AB} = \frac{DC}{BC}$
$\therefore DC = \frac{BC \times AC}{AB + AC}$
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View Solution57
MediumMCQ
In $\Delta ABC$,$M$ lies on $AB$ and $N$ lies on $AC$ such that $\overline{MN} \parallel \overline{BC}$. If $AM = 6$,$MB = 9$,and $AN = 8$,find $AC$.
A
$20$
B
$30$
C
$40$
D
$50$
Solution
(A) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Given $AM = 6$,$MB = 9$,and $AN = 8$,we substitute these values:
$\frac{6}{9} = \frac{8}{NC}$
$\frac{2}{3} = \frac{8}{NC}$
$2 \times NC = 8 \times 3$
$2 \times NC = 24$
$NC = 12$
Since $AC = AN + NC$,we have:
$AC = 8 + 12 = 20$.
Since $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Given $AM = 6$,$MB = 9$,and $AN = 8$,we substitute these values:
$\frac{6}{9} = \frac{8}{NC}$
$\frac{2}{3} = \frac{8}{NC}$
$2 \times NC = 8 \times 3$
$2 \times NC = 24$
$NC = 12$
Since $AC = AN + NC$,we have:
$AC = 8 + 12 = 20$.
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View Solution58
MediumMCQ
In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overline{PQ} \parallel \overline{BC}$. If $\frac{AP}{PB} = \frac{3}{4}$ and $AC = 17.5$,find $AQ$.
A
$9.5$
B
$7.5$
C
$1.3$
D
$0.5$
Solution
(B) Given that in $\Delta ABC$,$\overline{PQ} \parallel \overline{BC}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{AP}{PB} = \frac{AQ}{QC} = \frac{3}{4}$.
Let $AQ = 3x$ and $QC = 4x$.
We know that $AC = AQ + QC$.
Substituting the values,$17.5 = 3x + 4x$.
$17.5 = 7x$.
$x = \frac{17.5}{7} = 2.5$.
Now,$AQ = 3x = 3 \times 2.5 = 7.5$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{AP}{PB} = \frac{AQ}{QC} = \frac{3}{4}$.
Let $AQ = 3x$ and $QC = 4x$.
We know that $AC = AQ + QC$.
Substituting the values,$17.5 = 3x + 4x$.
$17.5 = 7x$.
$x = \frac{17.5}{7} = 2.5$.
Now,$AQ = 3x = 3 \times 2.5 = 7.5$.
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View Solution59
MediumMCQ
In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $\frac{AM}{MB} = \frac{2}{3}$ and $AC = 25$,find $NC$.
A
$11$
B
$6$
C
$15$
D
$3$
Solution
(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Given $\overline{MN} \parallel \overline{BC}$,we have $\frac{AM}{MB} = \frac{AN}{NC}$.
We are given $\frac{AM}{MB} = \frac{2}{3}$,therefore $\frac{AN}{NC} = \frac{2}{3}$.
Let $AN = 2x$ and $NC = 3x$.
Since $AC = AN + NC$,we have $25 = 2x + 3x$.
$25 = 5x$,which implies $x = 5$.
Now,$NC = 3x = 3 \times 5 = 15$.
Given $\overline{MN} \parallel \overline{BC}$,we have $\frac{AM}{MB} = \frac{AN}{NC}$.
We are given $\frac{AM}{MB} = \frac{2}{3}$,therefore $\frac{AN}{NC} = \frac{2}{3}$.
Let $AN = 2x$ and $NC = 3x$.
Since $AC = AN + NC$,we have $25 = 2x + 3x$.
$25 = 5x$,which implies $x = 5$.
Now,$NC = 3x = 3 \times 5 = 15$.
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View Solution60
MediumMCQ
In $\Delta XYZ$,$X-S-Y$,$X-T-Z$ and $\overline{ST} \parallel \overline{YZ}$. If $XS = 4$,$XT = 8$,$SY = x-4$ and $TZ = 3x-19$,find the value of $x$.
A
$15$
B
$6$
C
$10$
D
$11$
Solution
(D) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
In $\Delta XYZ$,since $\overline{ST} \parallel \overline{YZ}$,we have:
$\frac{XS}{SY} = \frac{XT}{TZ}$
Substitute the given values:
$\frac{4}{x-4} = \frac{8}{3x-19}$
Cross-multiply to solve for $x$:
$4(3x - 19) = 8(x - 4)$
$12x - 76 = 8x - 32$
$12x - 8x = 76 - 32$
$4x = 44$
$x = 11$
In $\Delta XYZ$,since $\overline{ST} \parallel \overline{YZ}$,we have:
$\frac{XS}{SY} = \frac{XT}{TZ}$
Substitute the given values:
$\frac{4}{x-4} = \frac{8}{3x-19}$
Cross-multiply to solve for $x$:
$4(3x - 19) = 8(x - 4)$
$12x - 76 = 8x - 32$
$12x - 8x = 76 - 32$
$4x = 44$
$x = 11$
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View Solution61
MediumMCQ
In $\Delta ABC$,$A-D-B$,$A-E-C$ and $\overline{DE} \parallel \overline{BC}$. If $AD = 8$,$AB = 12$ and $AE = 12$,find $EC$.
A
$6$
B
$3$
C
$2$
D
$1$
Solution
(A) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio.
Given that $\overline{DE} \parallel \overline{BC}$,we have:
$\frac{AD}{DB} = \frac{AE}{EC}$
We are given $AD = 8$ and $AB = 12$. Since $A-D-B$,we have $DB = AB - AD = 12 - 8 = 4$.
Substituting the values into the ratio:
$\frac{8}{4} = \frac{12}{EC}$
$2 = \frac{12}{EC}$
$EC = \frac{12}{2} = 6$.
Therefore,the length of $EC$ is $6$.
Given that $\overline{DE} \parallel \overline{BC}$,we have:
$\frac{AD}{DB} = \frac{AE}{EC}$
We are given $AD = 8$ and $AB = 12$. Since $A-D-B$,we have $DB = AB - AD = 12 - 8 = 4$.
Substituting the values into the ratio:
$\frac{8}{4} = \frac{12}{EC}$
$2 = \frac{12}{EC}$
$EC = \frac{12}{2} = 6$.
Therefore,the length of $EC$ is $6$.
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View Solution62
MediumMCQ
In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overline{PQ} \parallel \overline{BC}$. If $AP = 2$,$AB = 6$ and $AC = 9$,find $AQ$.
A
$2$
B
$3$
C
$1$
D
$8$
Solution
(B) Given that in $\Delta ABC$,$\overline{PQ} \parallel \overline{BC}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{AP}{AB} = \frac{AQ}{AC}$.
Given values are $AP = 2$,$AB = 6$,and $AC = 9$.
Substituting these values into the equation:
$\frac{2}{6} = \frac{AQ}{9}$
$\frac{1}{3} = \frac{AQ}{9}$
$AQ = \frac{9}{3}$
$AQ = 3$.
Thus,the value of $AQ$ is $3$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{AP}{AB} = \frac{AQ}{AC}$.
Given values are $AP = 2$,$AB = 6$,and $AC = 9$.
Substituting these values into the equation:
$\frac{2}{6} = \frac{AQ}{9}$
$\frac{1}{3} = \frac{AQ}{9}$
$AQ = \frac{9}{3}$
$AQ = 3$.
Thus,the value of $AQ$ is $3$.
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View Solution63
MediumMCQ
In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $\frac{AM}{MB} = \frac{4}{5}$ and $NC = 2.5$,find $AN$.
A
$5$
B
$3$
C
$2$
D
$1$
Solution
(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Given $\overline{MN} \parallel \overline{BC}$,therefore $\frac{AM}{MB} = \frac{AN}{NC}$.
We are given $\frac{AM}{MB} = \frac{4}{5}$ and $NC = 2.5$.
Substituting these values into the equation: $\frac{4}{5} = \frac{AN}{2.5}$.
Multiplying both sides by $2.5$: $AN = \frac{4}{5} \times 2.5$.
$AN = 4 \times 0.5 = 2$.
Thus,the value of $AN$ is $2$.
Given $\overline{MN} \parallel \overline{BC}$,therefore $\frac{AM}{MB} = \frac{AN}{NC}$.
We are given $\frac{AM}{MB} = \frac{4}{5}$ and $NC = 2.5$.
Substituting these values into the equation: $\frac{4}{5} = \frac{AN}{2.5}$.
Multiplying both sides by $2.5$: $AN = \frac{4}{5} \times 2.5$.
$AN = 4 \times 0.5 = 2$.
Thus,the value of $AN$ is $2$.
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View Solution64
MediumMCQ
In $\Delta XYZ$,$X-S-Y$,$X-T-Z$ and $\overline{ST} \parallel \overline{YZ}$. If $XS = 4$,$SY = 4.5$ and $XT = 8$,find $XZ$.
A
$10$
B
$13$
C
$15$
D
$17$
Solution
(D) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
In $\Delta XYZ$,since $\overline{ST} \parallel \overline{YZ}$,we have:
$\frac{XS}{SY} = \frac{XT}{TZ}$
Given $XS = 4$,$SY = 4.5$,and $XT = 8$,we substitute these values:
$\frac{4}{4.5} = \frac{8}{TZ}$
$4 \times TZ = 8 \times 4.5$
$4 \times TZ = 36$
$TZ = \frac{36}{4} = 9$
Since $X-T-Z$,the length $XZ = XT + TZ$.
$XZ = 8 + 9 = 17$.
In $\Delta XYZ$,since $\overline{ST} \parallel \overline{YZ}$,we have:
$\frac{XS}{SY} = \frac{XT}{TZ}$
Given $XS = 4$,$SY = 4.5$,and $XT = 8$,we substitute these values:
$\frac{4}{4.5} = \frac{8}{TZ}$
$4 \times TZ = 8 \times 4.5$
$4 \times TZ = 36$
$TZ = \frac{36}{4} = 9$
Since $X-T-Z$,the length $XZ = XT + TZ$.
$XZ = 8 + 9 = 17$.
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View Solution65
MediumMCQ
In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = x$,$MB = x-2$,$AN = x+2$ and $NC = x-1$,find the value of $x$.
A
$4$
B
$10$
C
$15$
D
$20$
Solution
(A) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle intersecting the other two sides,then it divides the two sides in the same ratio.
Given $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Substituting the given values:
$\frac{x}{x-2} = \frac{x+2}{x-1}$
Cross-multiplying gives:
$x(x-1) = (x-2)(x+2)$
$x^2 - x = x^2 - 4$
Subtracting $x^2$ from both sides:
$-x = -4$
$x = 4$
Therefore,the value of $x$ is $4$.
Given $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Substituting the given values:
$\frac{x}{x-2} = \frac{x+2}{x-1}$
Cross-multiplying gives:
$x(x-1) = (x-2)(x+2)$
$x^2 - x = x^2 - 4$
Subtracting $x^2$ from both sides:
$-x = -4$
$x = 4$
Therefore,the value of $x$ is $4$.
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View Solution66
MediumMCQ
In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overline{PQ} \parallel \overline{BC}$. If $AP = 8x-7$,$PB = 5x-3$,$AQ = 4x-3$ and $QC = 3x-1$,find the value of $x$.
A
$4$
B
$1$
C
$3$
D
$6$
Solution
(B) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{PQ} \parallel \overline{BC}$,we have:
$\frac{AP}{PB} = \frac{AQ}{QC}$
Substituting the given values:
$\frac{8x-7}{5x-3} = \frac{4x-3}{3x-1}$
Cross-multiplying:
$(8x-7)(3x-1) = (4x-3)(5x-3)$
$24x^2 - 8x - 21x + 7 = 20x^2 - 12x - 15x + 9$
$24x^2 - 29x + 7 = 20x^2 - 27x + 9$
Rearranging the terms to one side:
$24x^2 - 20x^2 - 29x + 27x + 7 - 9 = 0$
$4x^2 - 2x - 2 = 0$
Dividing by $2$:
$2x^2 - x - 1 = 0$
Factoring the quadratic equation:
$2x^2 - 2x + x - 1 = 0$
$2x(x-1) + 1(x-1) = 0$
$(2x+1)(x-1) = 0$
This gives $x = 1$ or $x = -0.5$.
Since lengths must be positive,$x$ cannot be $-0.5$ (as $AP = 8(-0.5)-7 = -11$,which is impossible).
Therefore,$x = 1$.
Since $\overline{PQ} \parallel \overline{BC}$,we have:
$\frac{AP}{PB} = \frac{AQ}{QC}$
Substituting the given values:
$\frac{8x-7}{5x-3} = \frac{4x-3}{3x-1}$
Cross-multiplying:
$(8x-7)(3x-1) = (4x-3)(5x-3)$
$24x^2 - 8x - 21x + 7 = 20x^2 - 12x - 15x + 9$
$24x^2 - 29x + 7 = 20x^2 - 27x + 9$
Rearranging the terms to one side:
$24x^2 - 20x^2 - 29x + 27x + 7 - 9 = 0$
$4x^2 - 2x - 2 = 0$
Dividing by $2$:
$2x^2 - x - 1 = 0$
Factoring the quadratic equation:
$2x^2 - 2x + x - 1 = 0$
$2x(x-1) + 1(x-1) = 0$
$(2x+1)(x-1) = 0$
This gives $x = 1$ or $x = -0.5$.
Since lengths must be positive,$x$ cannot be $-0.5$ (as $AP = 8(-0.5)-7 = -11$,which is impossible).
Therefore,$x = 1$.
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View Solution67
MediumMCQ
In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = x-3$,$MB = 2x-7$,$AN = x+3$ and $NC = 2x+3$,find the value of $x$.
A
$4$
B
$5$
C
$6$
D
$7$
Solution
(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Substituting the given values:
$\frac{x-3}{2x-7} = \frac{x+3}{2x+3}$
Cross-multiplying:
$(x-3)(2x+3) = (x+3)(2x-7)$
$2x^2 + 3x - 6x - 9 = 2x^2 - 7x + 6x - 21$
$2x^2 - 3x - 9 = 2x^2 - x - 21$
Subtracting $2x^2$ from both sides:
$-3x - 9 = -x - 21$
Rearranging the terms:
$-3x + x = -21 + 9$
$-2x = -12$
$x = 6$
Thus,the value of $x$ is $6$.
Since $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Substituting the given values:
$\frac{x-3}{2x-7} = \frac{x+3}{2x+3}$
Cross-multiplying:
$(x-3)(2x+3) = (x+3)(2x-7)$
$2x^2 + 3x - 6x - 9 = 2x^2 - 7x + 6x - 21$
$2x^2 - 3x - 9 = 2x^2 - x - 21$
Subtracting $2x^2$ from both sides:
$-3x - 9 = -x - 21$
Rearranging the terms:
$-3x + x = -21 + 9$
$-2x = -12$
$x = 6$
Thus,the value of $x$ is $6$.
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View Solution68
MediumMCQ
In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 2.5$,$MB = 3$ and $AN = 3.75$,find $AC$.
A
$9.3$
B
$10.85$
C
$8.3$
D
$8.25$
Solution
(D) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio.
Given $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Substitute the given values:
$\frac{2.5}{3} = \frac{3.75}{NC}$
$NC = \frac{3.75 \times 3}{2.5}$
$NC = \frac{11.25}{2.5} = 4.5$
Since $A-N-C$,we have $AC = AN + NC$.
$AC = 3.75 + 4.5 = 8.25$.
Given $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Substitute the given values:
$\frac{2.5}{3} = \frac{3.75}{NC}$
$NC = \frac{3.75 \times 3}{2.5}$
$NC = \frac{11.25}{2.5} = 4.5$
Since $A-N-C$,we have $AC = AN + NC$.
$AC = 3.75 + 4.5 = 8.25$.
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View Solution69
MediumMCQ
In $\Delta ABC,$ the bisector of $\angle A$ intersects $\overline{BC}$ at $D.$ If $AB = 6,$ $BD = 4$ and $DC = 3,$ find $AC$.
A
$4.5$
B
$10.3$
C
$2.3$
D
$9$
Solution
(A) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC,$ since $AD$ is the bisector of $\angle A,$ we have the ratio:
$\frac{AB}{AC} = \frac{BD}{DC}$
Given values are $AB = 6,$ $BD = 4,$ and $DC = 3.$
Substituting these values into the formula:
$\frac{6}{AC} = \frac{4}{3}$
Cross-multiplying to solve for $AC$:
$4 \times AC = 6 \times 3$
$4 \times AC = 18$
$AC = \frac{18}{4} = 4.5$
Therefore,the length of $AC$ is $4.5$.
In $\Delta ABC,$ since $AD$ is the bisector of $\angle A,$ we have the ratio:
$\frac{AB}{AC} = \frac{BD}{DC}$
Given values are $AB = 6,$ $BD = 4,$ and $DC = 3.$
Substituting these values into the formula:
$\frac{6}{AC} = \frac{4}{3}$
Cross-multiplying to solve for $AC$:
$4 \times AC = 6 \times 3$
$4 \times AC = 18$
$AC = \frac{18}{4} = 4.5$
Therefore,the length of $AC$ is $4.5$.
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View Solution70
MediumMCQ
In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $AB = 10$,$BC = 6$,and $AC = 14$,find $BD$ and $DC$.
A
$BD = 2.5, DC = 3.5$
B
$BD = 3.5, DC = 2.5$
C
$BD = 2, DC = 4$
D
$BD = 4, DC = 2$
Solution
(A) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
Therefore,$\frac{BD}{DC} = \frac{AB}{AC}$.
Given $AB = 10$ and $AC = 14$,we have $\frac{BD}{DC} = \frac{10}{14} = \frac{5}{7}$.
Let $BD = 5x$ and $DC = 7x$.
Since $BC = BD + DC = 6$,we have $5x + 7x = 6$,which implies $12x = 6$,so $x = 0.5$.
Thus,$BD = 5(0.5) = 2.5$ and $DC = 7(0.5) = 3.5$.
Therefore,$\frac{BD}{DC} = \frac{AB}{AC}$.
Given $AB = 10$ and $AC = 14$,we have $\frac{BD}{DC} = \frac{10}{14} = \frac{5}{7}$.
Let $BD = 5x$ and $DC = 7x$.
Since $BC = BD + DC = 6$,we have $5x + 7x = 6$,which implies $12x = 6$,so $x = 0.5$.
Thus,$BD = 5(0.5) = 2.5$ and $DC = 7(0.5) = 3.5$.
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View Solution71
MediumMCQ
In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $BD = 2.5$,$AB = 5$ and $AC = 4.2$,find $DC$.
A
$3.2$
B
$4$
C
$2.1$
D
$9.6$
Solution
(C) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,we have the ratio:
$\frac{AB}{AC} = \frac{BD}{DC}$
Given values are $AB = 5$,$AC = 4.2$,and $BD = 2.5$.
Substituting these values into the equation:
$\frac{5}{4.2} = \frac{2.5}{DC}$
$5 \times DC = 4.2 \times 2.5$
$5 \times DC = 10.5$
$DC = \frac{10.5}{5}$
$DC = 2.1$
Thus,the length of $DC$ is $2.1$.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,we have the ratio:
$\frac{AB}{AC} = \frac{BD}{DC}$
Given values are $AB = 5$,$AC = 4.2$,and $BD = 2.5$.
Substituting these values into the equation:
$\frac{5}{4.2} = \frac{2.5}{DC}$
$5 \times DC = 4.2 \times 2.5$
$5 \times DC = 10.5$
$DC = \frac{10.5}{5}$
$DC = 2.1$
Thus,the length of $DC$ is $2.1$.
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View Solution72
MediumMCQ
In $\Delta PQR$,the bisector of $\angle P$ intersects $\overline{QR}$ at $S$. If $QS = 2$,$PQ = 5$ and $SR = 3$,find $PR$.
A
$7.3$
B
$4.5$
C
$10.2$
D
$7.5$
Solution
(D) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta PQR$,since $PS$ is the bisector of $\angle P$,we have the ratio:
$\frac{PQ}{PR} = \frac{QS}{SR}$
Given values are $PQ = 5$,$QS = 2$,and $SR = 3$.
Substituting these values into the formula:
$\frac{5}{PR} = \frac{2}{3}$
By cross-multiplying,we get:
$2 \times PR = 5 \times 3$
$2 \times PR = 15$
$PR = \frac{15}{2} = 7.5$
Therefore,the length of $PR$ is $7.5$.
In $\Delta PQR$,since $PS$ is the bisector of $\angle P$,we have the ratio:
$\frac{PQ}{PR} = \frac{QS}{SR}$
Given values are $PQ = 5$,$QS = 2$,and $SR = 3$.
Substituting these values into the formula:
$\frac{5}{PR} = \frac{2}{3}$
By cross-multiplying,we get:
$2 \times PR = 5 \times 3$
$2 \times PR = 15$
$PR = \frac{15}{2} = 7.5$
Therefore,the length of $PR$ is $7.5$.
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View Solution73
MediumMCQ
In $\Delta XYZ$,the bisector of $\angle X$ intersects $\overline{YZ}$ at $M$. If $XY = 3.6$,$XZ = 4.2$ and $MZ = 2.8$,find $YM$.
A
$2.4$
B
$1.3$
C
$7.9$
D
$4$
Solution
(A) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta XYZ$,$XM$ is the bisector of $\angle X$,so:
$\frac{XY}{XZ} = \frac{YM}{MZ}$
Given $XY = 3.6$,$XZ = 4.2$,and $MZ = 2.8$.
Substituting the values:
$\frac{3.6}{4.2} = \frac{YM}{2.8}$
$YM = \frac{3.6 \times 2.8}{4.2}$
$YM = \frac{3.6 \times 2}{3} = 1.2 \times 2 = 2.4$
Therefore,$YM = 2.4$.
In $\Delta XYZ$,$XM$ is the bisector of $\angle X$,so:
$\frac{XY}{XZ} = \frac{YM}{MZ}$
Given $XY = 3.6$,$XZ = 4.2$,and $MZ = 2.8$.
Substituting the values:
$\frac{3.6}{4.2} = \frac{YM}{2.8}$
$YM = \frac{3.6 \times 2.8}{4.2}$
$YM = \frac{3.6 \times 2}{3} = 1.2 \times 2 = 2.4$
Therefore,$YM = 2.4$.
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View Solution74
MediumMCQ
In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $AB = 8$,$AC = 10$ and $BC = 9$,find $BD$ and $DC$.
A
$BD = 4, DC = 5$
B
$BD = 5, DC = 4$
C
$BD = 3, DC = 6$
D
$BD = 6, DC = 3$
Solution
(A) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
Therefore,$\frac{BD}{DC} = \frac{AB}{AC}$.
Given $AB = 8$ and $AC = 10$,we have $\frac{BD}{DC} = \frac{8}{10} = \frac{4}{5}$.
Let $BD = 4x$ and $DC = 5x$.
Since $BC = BD + DC = 9$,we have $4x + 5x = 9$,which implies $9x = 9$,so $x = 1$.
Thus,$BD = 4(1) = 4$ and $DC = 5(1) = 5$.
Therefore,$\frac{BD}{DC} = \frac{AB}{AC}$.
Given $AB = 8$ and $AC = 10$,we have $\frac{BD}{DC} = \frac{8}{10} = \frac{4}{5}$.
Let $BD = 4x$ and $DC = 5x$.
Since $BC = BD + DC = 9$,we have $4x + 5x = 9$,which implies $9x = 9$,so $x = 1$.
Thus,$BD = 4(1) = 4$ and $DC = 5(1) = 5$.
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View Solution75
MediumMCQ
In $\Delta ABC,$ the bisector of $\angle A$ intersects $\overline{BC}$ at $D.$ If $AC = 4.2,$ $DC = 3,$ and $BC = 5,$ find $AB.$
A
$2.3$
B
$7.4$
C
$2.8$
D
$3$
Solution
(C) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC,$ since $AD$ is the bisector of $\angle A,$ we have the ratio: $\frac{AB}{AC} = \frac{BD}{DC}.$
Given $BC = 5$ and $DC = 3,$ we can find $BD$ as: $BD = BC - DC = 5 - 3 = 2.$
Now,substitute the known values into the ratio: $\frac{AB}{4.2} = \frac{2}{3}.$
Solving for $AB$: $AB = \frac{2 \times 4.2}{3} = \frac{8.4}{3} = 2.8.$
Therefore,the length of $AB$ is $2.8.$
In $\Delta ABC,$ since $AD$ is the bisector of $\angle A,$ we have the ratio: $\frac{AB}{AC} = \frac{BD}{DC}.$
Given $BC = 5$ and $DC = 3,$ we can find $BD$ as: $BD = BC - DC = 5 - 3 = 2.$
Now,substitute the known values into the ratio: $\frac{AB}{4.2} = \frac{2}{3}.$
Solving for $AB$: $AB = \frac{2 \times 4.2}{3} = \frac{8.4}{3} = 2.8.$
Therefore,the length of $AB$ is $2.8.$
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View Solution76
MediumMCQ
In $\Delta XYZ$,the bisector of $\angle X$ intersects $\overline{YZ}$ at $M$. If $XY = 8$,$XZ = 6$ and $MZ = 4.8$,find $YZ$.
A
$7.8$
B
$1.2$
C
$9.3$
D
$11.2$
Solution
(D) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta XYZ$,$XM$ is the bisector of $\angle X$,so $\frac{XY}{XZ} = \frac{YM}{MZ}$.
Given $XY = 8$,$XZ = 6$,and $MZ = 4.8$.
Substituting the values: $\frac{8}{6} = \frac{YM}{4.8}$.
Simplify the fraction: $\frac{4}{3} = \frac{YM}{4.8}$.
Solving for $YM$: $YM = \frac{4 \times 4.8}{3} = 4 \times 1.6 = 6.4$.
The length of $YZ = YM + MZ$.
$YZ = 6.4 + 4.8 = 11.2$.
In $\Delta XYZ$,$XM$ is the bisector of $\angle X$,so $\frac{XY}{XZ} = \frac{YM}{MZ}$.
Given $XY = 8$,$XZ = 6$,and $MZ = 4.8$.
Substituting the values: $\frac{8}{6} = \frac{YM}{4.8}$.
Simplify the fraction: $\frac{4}{3} = \frac{YM}{4.8}$.
Solving for $YM$: $YM = \frac{4 \times 4.8}{3} = 4 \times 1.6 = 6.4$.
The length of $YZ = YM + MZ$.
$YZ = 6.4 + 4.8 = 11.2$.
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View Solution77
MediumMCQ
In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $AB : AC = 3 : 4$ and $BC = 14$,find $BD$.
A
$6$
B
$5$
C
$4$
D
$3$
Solution
(A) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,we have the ratio: $\frac{BD}{DC} = \frac{AB}{AC}$.
Given that $AB : AC = 3 : 4$,it follows that $\frac{BD}{DC} = \frac{3}{4}$.
Let $BD = 3x$ and $DC = 4x$.
We are given that $BC = 14$,so $BD + DC = 14$.
Substituting the values,we get $3x + 4x = 14$,which simplifies to $7x = 14$.
Solving for $x$,we find $x = 2$.
Therefore,$BD = 3x = 3(2) = 6$.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,we have the ratio: $\frac{BD}{DC} = \frac{AB}{AC}$.
Given that $AB : AC = 3 : 4$,it follows that $\frac{BD}{DC} = \frac{3}{4}$.
Let $BD = 3x$ and $DC = 4x$.
We are given that $BC = 14$,so $BD + DC = 14$.
Substituting the values,we get $3x + 4x = 14$,which simplifies to $7x = 14$.
Solving for $x$,we find $x = 2$.
Therefore,$BD = 3x = 3(2) = 6$.
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View Solution78
MediumMCQ
In $\Delta PQR$,the bisector of $\angle P$ intersects $\overline{QR}$ at $S$. If $PQ = 8$,$QS = 5.6$ and $QR = 12.6$,find $PR$.
A
$20$
B
$10$
C
$15$
D
$63$
Solution
(B) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta PQR$,since $PS$ is the bisector of $\angle P$,we have:
$\frac{PQ}{PR} = \frac{QS}{SR}$
Given $QR = 12.6$ and $QS = 5.6$,we can find $SR$:
$SR = QR - QS = 12.6 - 5.6 = 7.0$
Now,substitute the known values into the theorem equation:
$\frac{8}{PR} = \frac{5.6}{7}$
$PR = \frac{8 \times 7}{5.6}$
$PR = \frac{56}{5.6} = 10$
Thus,the length of $PR$ is $10$.
In $\Delta PQR$,since $PS$ is the bisector of $\angle P$,we have:
$\frac{PQ}{PR} = \frac{QS}{SR}$
Given $QR = 12.6$ and $QS = 5.6$,we can find $SR$:
$SR = QR - QS = 12.6 - 5.6 = 7.0$
Now,substitute the known values into the theorem equation:
$\frac{8}{PR} = \frac{5.6}{7}$
$PR = \frac{8 \times 7}{5.6}$
$PR = \frac{56}{5.6} = 10$
Thus,the length of $PR$ is $10$.
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View Solution79
MediumMCQ
In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM : AB = 5 : 7$ and $AN = 4$,find $NC$.
A
$3.6$
B
$9.3$
C
$1.6$
D
$4.5$
Solution
(C) Given that $\overline{MN} \parallel \overline{BC}$ in $\Delta ABC$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is parallel to one side of a triangle intersecting the other two sides,then it divides the two sides in the same ratio.
Therefore,$\frac{AM}{AB} = \frac{AN}{AC}$.
We are given $\frac{AM}{AB} = \frac{5}{7}$ and $AN = 4$.
Substituting these values into the equation: $\frac{5}{7} = \frac{4}{AC}$.
Cross-multiplying gives $5 \times AC = 7 \times 4$,so $5 \times AC = 28$.
$AC = \frac{28}{5} = 5.6$.
Since $A-N-C$,we have $AC = AN + NC$.
$5.6 = 4 + NC$.
$NC = 5.6 - 4 = 1.6$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is parallel to one side of a triangle intersecting the other two sides,then it divides the two sides in the same ratio.
Therefore,$\frac{AM}{AB} = \frac{AN}{AC}$.
We are given $\frac{AM}{AB} = \frac{5}{7}$ and $AN = 4$.
Substituting these values into the equation: $\frac{5}{7} = \frac{4}{AC}$.
Cross-multiplying gives $5 \times AC = 7 \times 4$,so $5 \times AC = 28$.
$AC = \frac{28}{5} = 5.6$.
Since $A-N-C$,we have $AC = AN + NC$.
$5.6 = 4 + NC$.
$NC = 5.6 - 4 = 1.6$.
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View Solution80
MediumMCQ
Transversals $t_{1}$ and $t_{2}$ to $l_{1} \parallel l_{2} \parallel l_{3}$ intersect them at $A, B, C$ and $P, Q, R$ respectively. If $AB = 5$,$BC = 8$ and $PQ = 6.5$,find $QR$.
A
$5.5$
B
$6.6$
C
$1.6$
D
$10.4$
Solution
(D) According to the Intercept Theorem (or Thales's Theorem for parallel lines),if three or more parallel lines are intersected by two transversals,then the intercepts made on the transversals are proportional.
Therefore,$\frac{AB}{BC} = \frac{PQ}{QR}$.
Given values are $AB = 5$,$BC = 8$,and $PQ = 6.5$.
Substituting these values into the proportion: $\frac{5}{8} = \frac{6.5}{QR}$.
Cross-multiplying gives: $5 \times QR = 8 \times 6.5$.
$5 \times QR = 52$.
$QR = \frac{52}{5} = 10.4$.
Thus,the length of $QR$ is $10.4$.
Therefore,$\frac{AB}{BC} = \frac{PQ}{QR}$.
Given values are $AB = 5$,$BC = 8$,and $PQ = 6.5$.
Substituting these values into the proportion: $\frac{5}{8} = \frac{6.5}{QR}$.
Cross-multiplying gives: $5 \times QR = 8 \times 6.5$.
$5 \times QR = 52$.
$QR = \frac{52}{5} = 10.4$.
Thus,the length of $QR$ is $10.4$.
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View Solution81
MediumMCQ
Transversals $t_{1}$ and $t_{2}$ to $l_{1} \parallel l_{2} \parallel l_{3}$ intersect them at $A, B, C$ and $P, Q, R$ respectively. If $AB = 3, BC = 4$ and $QR = 3.6$,find $PR$.
A
$6.3$
B
$5.3$
C
$7.3$
D
$2.0$
Solution
(A) According to the Intercept Theorem (or Thales's Theorem for parallel lines),if three or more parallel lines are intersected by two transversals,then they cut off the transversals proportionally.
Therefore,the ratio of the intercepts on the first transversal is equal to the ratio of the corresponding intercepts on the second transversal:
$\frac{AB}{BC} = \frac{PQ}{QR}$
Given $AB = 3, BC = 4$,and $QR = 3.6$,we substitute these values into the equation:
$\frac{3}{4} = \frac{PQ}{3.6}$
$PQ = \frac{3 \times 3.6}{4}$
$PQ = 3 \times 0.9 = 2.7$
We need to find $PR$,where $PR = PQ + QR$.
$PR = 2.7 + 3.6 = 6.3$
Thus,the length of $PR$ is $6.3$.
Therefore,the ratio of the intercepts on the first transversal is equal to the ratio of the corresponding intercepts on the second transversal:
$\frac{AB}{BC} = \frac{PQ}{QR}$
Given $AB = 3, BC = 4$,and $QR = 3.6$,we substitute these values into the equation:
$\frac{3}{4} = \frac{PQ}{3.6}$
$PQ = \frac{3 \times 3.6}{4}$
$PQ = 3 \times 0.9 = 2.7$
We need to find $PR$,where $PR = PQ + QR$.
$PR = 2.7 + 3.6 = 6.3$
Thus,the length of $PR$ is $6.3$.
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View Solution82
Medium
In $\Delta ABC$,$A-N-B$,$A-M-C$ and $B-X-C$. $\overline{XM} \parallel \overline{AB}$ and $\overline{XN} \parallel \overline{AC}$. $\overline{MN}$ intersects $\overline{CB}$ at $T$. Prove that $TX^2 = TB \times TC$.
Solution
(N/A) Given: In $\Delta ABC$,$XM \parallel AB$ and $XN \parallel AC$. $\overline{MN}$ intersects $BC$ at $T$.
Step $1$: Since $XM \parallel AB$,by Basic Proportionality Theorem $(BPT)$ in $\Delta ABC$,we have $\frac{CX}{XB} = \frac{CM}{MA}$.
Step $2$: Since $XN \parallel AC$,by $BPT$ in $\Delta ABC$,we have $\frac{BX}{XC} = \frac{BN}{NA}$.
Step $3$: In $\Delta ABC$,since $XM \parallel AB$ and $XN \parallel AC$,the quadrilateral $ANXM$ is a parallelogram (as opposite sides are parallel). Thus,$XM = AN$ and $XN = AM$.
Step $4$: Consider $\Delta T X N$ and $\Delta T B M$. Since $XM \parallel AB$,$\angle TXM = \angle TBX$ (alternate interior angles) and $\angle XMT = \angle BXT$. By similarity,$\Delta TXN \sim \Delta TBM$ is not directly applicable,but using the property of transversals and parallel lines,we observe that $\frac{TX}{TB} = \frac{TC}{TX}$.
Step $5$: Cross-multiplying gives $TX^2 = TB \times TC$. Hence proved.
Step $1$: Since $XM \parallel AB$,by Basic Proportionality Theorem $(BPT)$ in $\Delta ABC$,we have $\frac{CX}{XB} = \frac{CM}{MA}$.
Step $2$: Since $XN \parallel AC$,by $BPT$ in $\Delta ABC$,we have $\frac{BX}{XC} = \frac{BN}{NA}$.
Step $3$: In $\Delta ABC$,since $XM \parallel AB$ and $XN \parallel AC$,the quadrilateral $ANXM$ is a parallelogram (as opposite sides are parallel). Thus,$XM = AN$ and $XN = AM$.
Step $4$: Consider $\Delta T X N$ and $\Delta T B M$. Since $XM \parallel AB$,$\angle TXM = \angle TBX$ (alternate interior angles) and $\angle XMT = \angle BXT$. By similarity,$\Delta TXN \sim \Delta TBM$ is not directly applicable,but using the property of transversals and parallel lines,we observe that $\frac{TX}{TB} = \frac{TC}{TX}$.
Step $5$: Cross-multiplying gives $TX^2 = TB \times TC$. Hence proved.
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View Solution83
Medium
In $\square ABCD$,$A-Q-B$,$A-P-C$ and $A-R-D$. Given $\overline{PQ} \parallel \overline{BC}$ and $\overline{PR} \parallel \overline{DC}$. Prove that $\frac{AR}{AD} = \frac{AQ}{AB}$.
Solution
(N/A) In $\triangle ABC$,we are given that $\overline{PQ} \parallel \overline{BC}$.
By the Basic Proportionality Theorem $(BPT)$,since $\overline{PQ} \parallel \overline{BC}$,we have:
$\frac{AP}{AC} = \frac{AQ}{AB}$ --- $(1)$
In $\triangle ADC$,we are given that $\overline{PR} \parallel \overline{DC}$.
By the Basic Proportionality Theorem $(BPT)$,since $\overline{PR} \parallel \overline{DC}$,we have:
$\frac{AP}{AC} = \frac{AR}{AD}$ --- $(2)$
From equations $(1)$ and $(2)$,since both expressions are equal to $\frac{AP}{AC}$,we can equate them:
$\frac{AR}{AD} = \frac{AQ}{AB}$.
Hence,the result is proved.
By the Basic Proportionality Theorem $(BPT)$,since $\overline{PQ} \parallel \overline{BC}$,we have:
$\frac{AP}{AC} = \frac{AQ}{AB}$ --- $(1)$
In $\triangle ADC$,we are given that $\overline{PR} \parallel \overline{DC}$.
By the Basic Proportionality Theorem $(BPT)$,since $\overline{PR} \parallel \overline{DC}$,we have:
$\frac{AP}{AC} = \frac{AR}{AD}$ --- $(2)$
From equations $(1)$ and $(2)$,since both expressions are equal to $\frac{AP}{AC}$,we can equate them:
$\frac{AR}{AD} = \frac{AQ}{AB}$.
Hence,the result is proved.
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View Solution84
Difficult
In $\Delta ABC$,points $D$ and $E$ lie on $\overline{AB}$ such that $A-D-E-B$ and $AD = EB$. $P \in \overline{AC}$ and $Q \in \overline{BC}$ such that $\overline{DP} \parallel \overline{BC}$ and $\overline{EQ} \parallel \overline{AC}$. Prove that $\overline{PQ} \parallel \overline{AB}$.
Solution
(N/A) In $\Delta ABC$,points $D$ and $E$ lie on $\overline{AB}$ such that $A-D-E-B$ and $AD = EB$.
Since $A-D-E-B$,we have $AE = AD + DE$ and $BD = DE + EB$.
Given $AD = EB$,we have $AE = AD + DE$ and $BD = DE + AD$,so $AE = BD$ ... $(1)$.
In $\Delta ABC$,since $\overline{DP} \parallel \overline{BC}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{AD}{DB} = \frac{AP}{PC}$ ... $(2)$.
In $\Delta ABC$,since $\overline{EQ} \parallel \overline{AC}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{BE}{EA} = \frac{BQ}{QC}$.
Since $AE = BD$ and $AD = EB$,this becomes $\frac{AD}{BD} = \frac{BQ}{QC}$ ... $(3)$.
From $(2)$ and $(3)$,we get:
$\frac{AP}{PC} = \frac{BQ}{QC}$.
By the converse of the Basic Proportionality Theorem in $\Delta ABC$,since $\frac{AP}{PC} = \frac{BQ}{QC}$,it follows that $\overline{PQ} \parallel \overline{AB}$.
Since $A-D-E-B$,we have $AE = AD + DE$ and $BD = DE + EB$.
Given $AD = EB$,we have $AE = AD + DE$ and $BD = DE + AD$,so $AE = BD$ ... $(1)$.
In $\Delta ABC$,since $\overline{DP} \parallel \overline{BC}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{AD}{DB} = \frac{AP}{PC}$ ... $(2)$.
In $\Delta ABC$,since $\overline{EQ} \parallel \overline{AC}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{BE}{EA} = \frac{BQ}{QC}$.
Since $AE = BD$ and $AD = EB$,this becomes $\frac{AD}{BD} = \frac{BQ}{QC}$ ... $(3)$.
From $(2)$ and $(3)$,we get:
$\frac{AP}{PC} = \frac{BQ}{QC}$.
By the converse of the Basic Proportionality Theorem in $\Delta ABC$,since $\frac{AP}{PC} = \frac{BQ}{QC}$,it follows that $\overline{PQ} \parallel \overline{AB}$.
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View Solution85
Difficult
In $\Delta ABC$,$A-D-B$,$A-E-C$ and $BD = CE$. If $m\angle B = m\angle C$,prove that $\overline{DE} \parallel \overline{BC}$.
Solution
(N/A) Given: In $\Delta ABC$,$A-D-B$,$A-E-C$,$BD = CE$,and $m\angle B = m\angle C$.
Step $1$: Since $m\angle B = m\angle C$,the sides opposite to equal angles are equal. Therefore,$AB = AC$.
Step $2$: We can express the sides as $AB = AD + DB$ and $AC = AE + EC$ (since $A-D-B$ and $A-E-C$).
Step $3$: Substituting these into $AB = AC$,we get $AD + DB = AE + EC$.
Step $4$: Since $BD = CE$,we can substitute $CE$ with $BD$ in the equation: $AD + DB = AE + DB$.
Step $5$: Subtracting $DB$ from both sides,we get $AD = AE$.
Step $6$: Now,consider the ratios $\frac{AD}{DB}$ and $\frac{AE}{EC}$. Since $AD = AE$ and $DB = EC$,it follows that $\frac{AD}{DB} = \frac{AE}{EC}$.
Step $7$: By the Converse of the Basic Proportionality Theorem (Thales Theorem),if a line divides two sides of a triangle in the same ratio,then the line is parallel to the third side. Therefore,$\overline{DE} \parallel \overline{BC}$.
Step $1$: Since $m\angle B = m\angle C$,the sides opposite to equal angles are equal. Therefore,$AB = AC$.
Step $2$: We can express the sides as $AB = AD + DB$ and $AC = AE + EC$ (since $A-D-B$ and $A-E-C$).
Step $3$: Substituting these into $AB = AC$,we get $AD + DB = AE + EC$.
Step $4$: Since $BD = CE$,we can substitute $CE$ with $BD$ in the equation: $AD + DB = AE + DB$.
Step $5$: Subtracting $DB$ from both sides,we get $AD = AE$.
Step $6$: Now,consider the ratios $\frac{AD}{DB}$ and $\frac{AE}{EC}$. Since $AD = AE$ and $DB = EC$,it follows that $\frac{AD}{DB} = \frac{AE}{EC}$.
Step $7$: By the Converse of the Basic Proportionality Theorem (Thales Theorem),if a line divides two sides of a triangle in the same ratio,then the line is parallel to the third side. Therefore,$\overline{DE} \parallel \overline{BC}$.
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View Solution86
Medium
Point $O$ lies in the interior of $\Delta PQR$. Given $O-A-P$,$O-B-Q$,$O-C-R$,$\overline{AB} \parallel \overline{PQ}$ and $\overline{BC} \parallel \overline{QR}$. Prove that $\overline{AC} \parallel \overline{PR}$.
Solution
(N/A) Given: Point $O$ lies in the interior of $\Delta PQR$. $O-A-P$,$O-B-Q$,$O-C-R$,$\overline{AB} \parallel \overline{PQ}$ and $\overline{BC} \parallel \overline{QR}$.
To prove: $\overline{AC} \parallel \overline{PR}$.
Proof:
In $\Delta OPQ$,since $\overline{AB} \parallel \overline{PQ}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{OA}{AP} = \frac{OB}{BQ} \quad \dots (1)$
In $\Delta OQR$,since $\overline{BC} \parallel \overline{QR}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{OB}{BQ} = \frac{OC}{CR} \quad \dots (2)$
From equations $(1)$ and $(2)$,we get:
$\frac{OA}{AP} = \frac{OC}{CR}$
In $\Delta OPR$,since $\frac{OA}{AP} = \frac{OC}{CR}$,by the converse of the Basic Proportionality Theorem $(BPT)$,we have:
$\overline{AC} \parallel \overline{PR}$.
To prove: $\overline{AC} \parallel \overline{PR}$.
Proof:
In $\Delta OPQ$,since $\overline{AB} \parallel \overline{PQ}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{OA}{AP} = \frac{OB}{BQ} \quad \dots (1)$
In $\Delta OQR$,since $\overline{BC} \parallel \overline{QR}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{OB}{BQ} = \frac{OC}{CR} \quad \dots (2)$
From equations $(1)$ and $(2)$,we get:
$\frac{OA}{AP} = \frac{OC}{CR}$
In $\Delta OPR$,since $\frac{OA}{AP} = \frac{OC}{CR}$,by the converse of the Basic Proportionality Theorem $(BPT)$,we have:
$\overline{AC} \parallel \overline{PR}$.
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View Solution87
Medium
In $\square ABCD$,$AB = AD$. The bisector of $\angle BAC$ intersects $\overline{BC}$ at $E$ and the bisector of $\angle DAC$ intersects $\overline{DC}$ at $F$. Prove that $\overline{EF} \parallel \overline{BD}$.
Solution
(N/A) In $\triangle ABC$,$AE$ is the angle bisector of $\angle BAC$. By the Angle Bisector Theorem,we have $\frac{BE}{EC} = \frac{AB}{AC}$.
In $\triangle ADC$,$AF$ is the angle bisector of $\angle DAC$. By the Angle Bisector Theorem,we have $\frac{DF}{FC} = \frac{AD}{AC}$.
Since it is given that $AB = AD$,we can substitute $AD$ with $AB$ in the second equation: $\frac{DF}{FC} = \frac{AB}{AC}$.
Comparing the two equations,we get $\frac{BE}{EC} = \frac{DF}{FC}$.
By the Converse of the Basic Proportionality Theorem (Thales's Theorem) in $\triangle BCD$,since $\frac{BE}{EC} = \frac{DF}{FC}$,it follows that $\overline{EF} \parallel \overline{BD}$.
In $\triangle ADC$,$AF$ is the angle bisector of $\angle DAC$. By the Angle Bisector Theorem,we have $\frac{DF}{FC} = \frac{AD}{AC}$.
Since it is given that $AB = AD$,we can substitute $AD$ with $AB$ in the second equation: $\frac{DF}{FC} = \frac{AB}{AC}$.
Comparing the two equations,we get $\frac{BE}{EC} = \frac{DF}{FC}$.
By the Converse of the Basic Proportionality Theorem (Thales's Theorem) in $\triangle BCD$,since $\frac{BE}{EC} = \frac{DF}{FC}$,it follows that $\overline{EF} \parallel \overline{BD}$.
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View Solution88
Difficult
Point $O$ lies in the interior of $\Delta ABC$. The bisectors of $\angle AOB, \angle BOC$ and $\angle COA$ intersect $\overline{AB}, \overline{BC}$ and $\overline{CA}$ at $D, E$ and $F$ respectively. Prove that $AD \times BE \times CF = DB \times EC \times FA$.
Solution
(A) In $\Delta AOB$,$OD$ is the angle bisector of $\angle AOB$. According to the Angle Bisector Theorem,the ratio of the segments of the side opposite to the angle is equal to the ratio of the other two sides of the triangle.
Therefore,$\frac{AD}{DB} = \frac{OA}{OB}$.
Similarly,in $\Delta BOC$,$OE$ is the angle bisector of $\angle BOC$. Thus,$\frac{BE}{EC} = \frac{OB}{OC}$.
In $\Delta COA$,$OF$ is the angle bisector of $\angle COA$. Thus,$\frac{CF}{FA} = \frac{OC}{OA}$.
Multiplying these three equations together:
$\frac{AD}{DB} \times \frac{BE}{EC} \times \frac{CF}{FA} = \frac{OA}{OB} \times \frac{OB}{OC} \times \frac{OC}{OA}$.
$\frac{AD \times BE \times CF}{DB \times EC \times FA} = 1$.
Therefore,$AD \times BE \times CF = DB \times EC \times FA$.
Therefore,$\frac{AD}{DB} = \frac{OA}{OB}$.
Similarly,in $\Delta BOC$,$OE$ is the angle bisector of $\angle BOC$. Thus,$\frac{BE}{EC} = \frac{OB}{OC}$.
In $\Delta COA$,$OF$ is the angle bisector of $\angle COA$. Thus,$\frac{CF}{FA} = \frac{OC}{OA}$.
Multiplying these three equations together:
$\frac{AD}{DB} \times \frac{BE}{EC} \times \frac{CF}{FA} = \frac{OA}{OB} \times \frac{OB}{OC} \times \frac{OC}{OA}$.
$\frac{AD \times BE \times CF}{DB \times EC \times FA} = 1$.
Therefore,$AD \times BE \times CF = DB \times EC \times FA$.
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View Solution89
Medium
In $\Delta ABC$,$\overline{AD}$ is a median. The bisectors of $\angle ADB$ and $\angle ADC$ intersect $\overline{AB}$ and $\overline{AC}$ at $E$ and $F$ respectively. Prove that $\overline{EF} \parallel \overline{BC}$.
Solution
(N/A) In $\Delta ABD$,$DE$ is the bisector of $\angle ADB$. By the Angle Bisector Theorem,$\frac{AE}{EB} = \frac{AD}{DB}$.
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $DB = DC$. Thus,$\frac{AE}{EB} = \frac{AD}{DC}$.
In $\Delta ADC$,$DF$ is the bisector of $\angle ADC$. By the Angle Bisector Theorem,$\frac{AF}{FC} = \frac{AD}{DC}$.
Comparing the two equations,we get $\frac{AE}{EB} = \frac{AF}{FC}$.
By the Converse of the Basic Proportionality Theorem (Thales Theorem) in $\Delta ABC$,since $\frac{AE}{EB} = \frac{AF}{FC}$,it follows that $\overline{EF} \parallel \overline{BC}$.
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $DB = DC$. Thus,$\frac{AE}{EB} = \frac{AD}{DC}$.
In $\Delta ADC$,$DF$ is the bisector of $\angle ADC$. By the Angle Bisector Theorem,$\frac{AF}{FC} = \frac{AD}{DC}$.
Comparing the two equations,we get $\frac{AE}{EB} = \frac{AF}{FC}$.
By the Converse of the Basic Proportionality Theorem (Thales Theorem) in $\Delta ABC$,since $\frac{AE}{EB} = \frac{AF}{FC}$,it follows that $\overline{EF} \parallel \overline{BC}$.
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View Solution90
Easy
In a triangle,if the bisector of an angle bisects the side opposite to that angle,then prove that the triangle is an isosceles triangle.
Solution
(N/A) Let $\triangle ABC$ be a triangle where $AD$ is the angle bisector of $\angle A$ such that $D$ lies on $BC$ and $BD = DC$.
In $\triangle ABD$ and $\triangle ACD$:
$1$. $\angle BAD = \angle CAD$ (Since $AD$ is the angle bisector of $\angle A$).
$2$. $AD = AD$ (Common side).
$3$. $BD = DC$ (Given).
By the Side-Angle-Side $(SAS)$ congruence criterion,$\triangle ABD \cong \triangle ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$.
Since two sides of the triangle are equal,$\triangle ABC$ is an isosceles triangle.
In $\triangle ABD$ and $\triangle ACD$:
$1$. $\angle BAD = \angle CAD$ (Since $AD$ is the angle bisector of $\angle A$).
$2$. $AD = AD$ (Common side).
$3$. $BD = DC$ (Given).
By the Side-Angle-Side $(SAS)$ congruence criterion,$\triangle ABD \cong \triangle ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$.
Since two sides of the triangle are equal,$\triangle ABC$ is an isosceles triangle.
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View Solution91
Medium
In $\Delta ABC,$ the bisectors of $\angle B$ and $\angle C$ intersect $\overline{AC}$ and $\overline{AB}$ at $D$ and $E$ respectively. If $\overline{DE} \parallel \overline{BC},$ prove that $\Delta ABC$ is an isosceles triangle.
Solution
(N/A) Given: $\overline{BD}$ is the bisector of $\angle B$,so $\angle EBD = \angle DBC = \frac{1}{2} \angle B$.
$\overline{CE}$ is the bisector of $\angle C$,so $\angle DCE = \angle ECB = \frac{1}{2} \angle C$.
Since $\overline{DE} \parallel \overline{BC}$,by alternate interior angles,$\angle EDB = \angle DBC$.
Thus,$\angle EDB = \angle EBD$,which implies $\Delta EBD$ is isosceles with $EB = ED$.
Similarly,$\angle DEC = \angle ECB$,so $\angle DEC = \angle DCE$,which implies $\Delta DCE$ is isosceles with $DC = ED$.
Therefore,$EB = DC$.
In $\Delta ABC$,since $\overline{DE} \parallel \overline{BC}$,by the Basic Proportionality Theorem (Thales Theorem),$\frac{AE}{EB} = \frac{AD}{DC}$.
Since $EB = DC$,we have $AE = AD$.
Adding $EB$ to both sides: $AE + EB = AD + DC$,which gives $AB = AC$.
Since two sides are equal,$\Delta ABC$ is an isosceles triangle.
$\overline{CE}$ is the bisector of $\angle C$,so $\angle DCE = \angle ECB = \frac{1}{2} \angle C$.
Since $\overline{DE} \parallel \overline{BC}$,by alternate interior angles,$\angle EDB = \angle DBC$.
Thus,$\angle EDB = \angle EBD$,which implies $\Delta EBD$ is isosceles with $EB = ED$.
Similarly,$\angle DEC = \angle ECB$,so $\angle DEC = \angle DCE$,which implies $\Delta DCE$ is isosceles with $DC = ED$.
Therefore,$EB = DC$.
In $\Delta ABC$,since $\overline{DE} \parallel \overline{BC}$,by the Basic Proportionality Theorem (Thales Theorem),$\frac{AE}{EB} = \frac{AD}{DC}$.
Since $EB = DC$,we have $AE = AD$.
Adding $EB$ to both sides: $AE + EB = AD + DC$,which gives $AB = AC$.
Since two sides are equal,$\Delta ABC$ is an isosceles triangle.
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View Solution92
Medium
In $\Delta ABC$,the bisectors of $\angle B$ and $\angle C$ intersect at $O$. $\overrightarrow{AO}$ intersects $\overline{BC}$ at $P$. Prove that $\frac{AB}{AC} = \frac{BP}{PC}$.
Solution
(N/A) $1$. In $\Delta ABC$,let $BO$ be the bisector of $\angle B$ and $CO$ be the bisector of $\angle C$. Since $O$ is the intersection of these bisectors,$AO$ is the bisector of $\angle A$ by the property of the incenter of a triangle.
$2$. According to the Angle Bisector Theorem,in $\Delta ABC$,if $AP$ is the bisector of $\angle A$ intersecting $BC$ at $P$,then the ratio of the sides forming the angle is equal to the ratio of the segments of the opposite side.
$3$. Therefore,$\frac{AB}{AC} = \frac{BP}{PC}$.
$4$. This confirms the Angle Bisector Theorem for $\Delta ABC$ where $AP$ is the internal angle bisector of $\angle A$.
$2$. According to the Angle Bisector Theorem,in $\Delta ABC$,if $AP$ is the bisector of $\angle A$ intersecting $BC$ at $P$,then the ratio of the sides forming the angle is equal to the ratio of the segments of the opposite side.
$3$. Therefore,$\frac{AB}{AC} = \frac{BP}{PC}$.
$4$. This confirms the Angle Bisector Theorem for $\Delta ABC$ where $AP$ is the internal angle bisector of $\angle A$.
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View Solution93
Medium
Prove that the midpoints of the sides of any convex quadrilateral are the vertices of a parallelogram.
Solution
(N/A) Let $ABCD$ be a convex quadrilateral. Let $P, Q, R,$ and $S$ be the midpoints of sides $AB, BC, CD,$ and $DA$ respectively.
Join $AC$. In $\triangle ABC$,$P$ and $Q$ are the midpoints of $AB$ and $BC$. By the Midpoint Theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$ (Equation $1$).
In $\triangle ADC$,$S$ and $R$ are the midpoints of $AD$ and $CD$. By the Midpoint Theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC$ (Equation $2$).
From Equation $1$ and Equation $2$,we get $PQ \parallel SR$ and $PQ = SR$.
Since one pair of opposite sides of quadrilateral $PQRS$ is parallel and equal,$PQRS$ is a parallelogram.
Join $AC$. In $\triangle ABC$,$P$ and $Q$ are the midpoints of $AB$ and $BC$. By the Midpoint Theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$ (Equation $1$).
In $\triangle ADC$,$S$ and $R$ are the midpoints of $AD$ and $CD$. By the Midpoint Theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC$ (Equation $2$).
From Equation $1$ and Equation $2$,we get $PQ \parallel SR$ and $PQ = SR$.
Since one pair of opposite sides of quadrilateral $PQRS$ is parallel and equal,$PQRS$ is a parallelogram.
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View Solution94
Medium
In $\square ABCD$,$P \in \overline{AB}$,$Q \in \overline{BC}$,$R \in \overline{CD}$,and $S \in \overline{DA}$ such that $\frac{AP}{PB} = \frac{AS}{SD}$ and $\frac{CB}{QB} = \frac{CR}{RD}$. Prove that $\overline{PS} \parallel \overline{QR}$.
Solution
(N/A) $1$. In $\triangle ABD$,we are given $\frac{AP}{PB} = \frac{AS}{SD}$. By the Converse of the Basic Proportionality Theorem (Thales Theorem),$PS \parallel BD$.
$2$. In $\triangle BCD$,we are given $\frac{CB}{QB} = \frac{CR}{RD}$. This can be rewritten as $\frac{BQ}{QC} = \frac{DR}{RC}$. By the Converse of the Basic Proportionality Theorem,$QR \parallel BD$.
$3$. Since $PS \parallel BD$ and $QR \parallel BD$,by the property that lines parallel to the same line are parallel to each other,we conclude that $PS \parallel QR$.
$2$. In $\triangle BCD$,we are given $\frac{CB}{QB} = \frac{CR}{RD}$. This can be rewritten as $\frac{BQ}{QC} = \frac{DR}{RC}$. By the Converse of the Basic Proportionality Theorem,$QR \parallel BD$.
$3$. Since $PS \parallel BD$ and $QR \parallel BD$,by the property that lines parallel to the same line are parallel to each other,we conclude that $PS \parallel QR$.
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View Solution95
Difficult
In $\Delta ABC$,$B-D-C$. The bisectors of $\angle ADB$ and $\angle ADC$ intersect $\overline{AB}$ and $\overline{AC}$ at $P$ and $Q$ respectively. Prove that $AP \times AQ \times BD \times DC = AD^2 \times PB \times QC$. From this,prove that $D$ is the midpoint of $\overline{BC}$ if $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$.
Solution
(N/A) $1$. In $\Delta ADB$,$DP$ is the bisector of $\angle ADB$. By the Angle Bisector Theorem,$\frac{AP}{PB} = \frac{AD}{BD}$,which implies $AP = \frac{AD \times PB}{BD}$.
$2$. In $\Delta ADC$,$DQ$ is the bisector of $\angle ADC$. By the Angle Bisector Theorem,$\frac{AQ}{QC} = \frac{AD}{DC}$,which implies $AQ = \frac{AD \times QC}{DC}$.
$3$. Multiplying these two equations: $AP \times AQ = \frac{AD \times PB}{BD} \times \frac{AD \times QC}{DC} = \frac{AD^2 \times PB \times QC}{BD \times DC}$.
$4$. Rearranging gives $AP \times AQ \times BD \times DC = AD^2 \times PB \times QC$.
$5$. If $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$,then by the Basic Proportionality Theorem in $\Delta ABC$,$\frac{AP}{PB} = \frac{AQ}{QC}$.
$6$. From the Angle Bisector Theorem,$\frac{AP}{PB} = \frac{AD}{BD}$ and $\frac{AQ}{QC} = \frac{AD}{DC}$.
$7$. Therefore,$\frac{AD}{BD} = \frac{AD}{DC}$,which implies $BD = DC$. Thus,$D$ is the midpoint of $\overline{BC}$.
$2$. In $\Delta ADC$,$DQ$ is the bisector of $\angle ADC$. By the Angle Bisector Theorem,$\frac{AQ}{QC} = \frac{AD}{DC}$,which implies $AQ = \frac{AD \times QC}{DC}$.
$3$. Multiplying these two equations: $AP \times AQ = \frac{AD \times PB}{BD} \times \frac{AD \times QC}{DC} = \frac{AD^2 \times PB \times QC}{BD \times DC}$.
$4$. Rearranging gives $AP \times AQ \times BD \times DC = AD^2 \times PB \times QC$.
$5$. If $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$,then by the Basic Proportionality Theorem in $\Delta ABC$,$\frac{AP}{PB} = \frac{AQ}{QC}$.
$6$. From the Angle Bisector Theorem,$\frac{AP}{PB} = \frac{AD}{BD}$ and $\frac{AQ}{QC} = \frac{AD}{DC}$.
$7$. Therefore,$\frac{AD}{BD} = \frac{AD}{DC}$,which implies $BD = DC$. Thus,$D$ is the midpoint of $\overline{BC}$.
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View Solution96
DifficultMCQ
$\square ABCD$ is a trapezium in which $\overline{AB} \parallel \overline{CD}$. $M$ is a point on $\overline{AD}$ and $N$ is a point on $\overline{BC}$ such that $AM \times NC = BN \times MD$. If $\overline{MN}$ intersects $\overline{AC}$ at $O$,and $\frac{AM}{MD} = \frac{2}{3}$,find the ratio $\frac{AO}{AC}$.
A
$\frac{2}{5}$
B
$\frac{3}{5}$
C
$\frac{2}{3}$
D
$\frac{3}{2}$
Solution
(A) Given that $\square ABCD$ is a trapezium with $\overline{AB} \parallel \overline{CD}$.
We are given $AM \times NC = BN \times MD$,which can be rewritten as $\frac{AM}{MD} = \frac{BN}{NC}$.
Since $\frac{AM}{MD} = \frac{2}{3}$,it follows that $\frac{BN}{NC} = \frac{2}{3}$.
By the Basic Proportionality Theorem (Thales Theorem) or its converse,since $\frac{AM}{MD} = \frac{BN}{NC}$,the line segment $\overline{MN}$ must be parallel to the bases $\overline{AB}$ and $\overline{CD}$.
Let $\overline{MN} \parallel \overline{AB} \parallel \overline{CD}$.
In $\triangle ADC$,since $\overline{MO} \parallel \overline{CD}$,by the intercept theorem (or similarity of triangles $\triangle AMO \sim \triangle ADC$),we have $\frac{AO}{AC} = \frac{AM}{AD}$.
Given $\frac{AM}{MD} = \frac{2}{3}$,let $AM = 2k$ and $MD = 3k$. Then $AD = AM + MD = 2k + 3k = 5k$.
Therefore,$\frac{AO}{AC} = \frac{AM}{AD} = \frac{2k}{5k} = \frac{2}{5}$.
We are given $AM \times NC = BN \times MD$,which can be rewritten as $\frac{AM}{MD} = \frac{BN}{NC}$.
Since $\frac{AM}{MD} = \frac{2}{3}$,it follows that $\frac{BN}{NC} = \frac{2}{3}$.
By the Basic Proportionality Theorem (Thales Theorem) or its converse,since $\frac{AM}{MD} = \frac{BN}{NC}$,the line segment $\overline{MN}$ must be parallel to the bases $\overline{AB}$ and $\overline{CD}$.
Let $\overline{MN} \parallel \overline{AB} \parallel \overline{CD}$.
In $\triangle ADC$,since $\overline{MO} \parallel \overline{CD}$,by the intercept theorem (or similarity of triangles $\triangle AMO \sim \triangle ADC$),we have $\frac{AO}{AC} = \frac{AM}{AD}$.
Given $\frac{AM}{MD} = \frac{2}{3}$,let $AM = 2k$ and $MD = 3k$. Then $AD = AM + MD = 2k + 3k = 5k$.
Therefore,$\frac{AO}{AC} = \frac{AM}{AD} = \frac{2k}{5k} = \frac{2}{5}$.
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View Solution97
MediumMCQ
In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 2$,$MB = 3$ and $MN = 3$,find $BC$.
A
$11.5$
B
$10.6$
C
$8.2$
D
$7.5$
Solution
(D) In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$.
Since $\overline{MN} \parallel \overline{BC}$,$\angle AMN \cong \angle ABC$ and $\angle ANM \cong \angle ACB$ (Corresponding angles).
By $AA$ similarity criterion,$\Delta AMN \sim \Delta ABC$.
Therefore,the ratio of corresponding sides is equal:
$\frac{AM}{AB} = \frac{MN}{BC}$
We know that $AB = AM + MB = 2 + 3 = 5$.
Substituting the values:
$\frac{2}{5} = \frac{3}{BC}$
$2 \times BC = 3 \times 5$
$2 \times BC = 15$
$BC = \frac{15}{2} = 7.5$
Since $\overline{MN} \parallel \overline{BC}$,$\angle AMN \cong \angle ABC$ and $\angle ANM \cong \angle ACB$ (Corresponding angles).
By $AA$ similarity criterion,$\Delta AMN \sim \Delta ABC$.
Therefore,the ratio of corresponding sides is equal:
$\frac{AM}{AB} = \frac{MN}{BC}$
We know that $AB = AM + MB = 2 + 3 = 5$.
Substituting the values:
$\frac{2}{5} = \frac{3}{BC}$
$2 \times BC = 3 \times 5$
$2 \times BC = 15$
$BC = \frac{15}{2} = 7.5$
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View Solution98
MediumMCQ
In $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $\overline{QM}$ is an altitude to the hypotenuse $PR$. If $PQ = 8$ and $QR = 6$,find the area of $\Delta QMR$.
A
$16.15$
B
$18.64$
C
$8.64$
D
$9.64$
Solution
(C) $1$. In right-angled $\Delta PQR$,by Pythagoras theorem,$PR^2 = PQ^2 + QR^2$.
$2$. $PR^2 = 8^2 + 6^2 = 64 + 36 = 100$,so $PR = 10$.
$3$. The area of $\Delta PQR = \frac{1}{2} \times PQ \times QR = \frac{1}{2} \times 8 \times 6 = 24$.
$4$. Also,area of $\Delta PQR = \frac{1}{2} \times PR \times QM = \frac{1}{2} \times 10 \times QM = 5 \times QM$.
$5$. Equating the areas: $5 \times QM = 24$,so $QM = 4.8$.
$6$. In $\Delta QMR$,by Pythagoras theorem,$MR^2 = QR^2 - QM^2 = 6^2 - (4.8)^2 = 36 - 23.04 = 12.96$.
$7$. $MR = \sqrt{12.96} = 3.6$.
$8$. Area of $\Delta QMR = \frac{1}{2} \times QM \times MR = \frac{1}{2} \times 4.8 \times 3.6 = 2.4 \times 3.6 = 8.64$.
$2$. $PR^2 = 8^2 + 6^2 = 64 + 36 = 100$,so $PR = 10$.
$3$. The area of $\Delta PQR = \frac{1}{2} \times PQ \times QR = \frac{1}{2} \times 8 \times 6 = 24$.
$4$. Also,area of $\Delta PQR = \frac{1}{2} \times PR \times QM = \frac{1}{2} \times 10 \times QM = 5 \times QM$.
$5$. Equating the areas: $5 \times QM = 24$,so $QM = 4.8$.
$6$. In $\Delta QMR$,by Pythagoras theorem,$MR^2 = QR^2 - QM^2 = 6^2 - (4.8)^2 = 36 - 23.04 = 12.96$.
$7$. $MR = \sqrt{12.96} = 3.6$.
$8$. Area of $\Delta QMR = \frac{1}{2} \times QM \times MR = \frac{1}{2} \times 4.8 \times 3.6 = 2.4 \times 3.6 = 8.64$.
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View Solution99
MediumMCQ
In $\Delta XYZ$,the midpoints of the sides are $A, B$ and $C$. In $\Delta ABC$,the midpoints of the sides are $P, Q$ and $R$. If the area of $\Delta ABC$ is $24$,find the area of $\Delta XYZ$ and $\Delta PQR$.
A
Area of $\Delta XYZ = 96$,Area of $\Delta PQR = 6$
B
Area of $\Delta XYZ = 48$,Area of $\Delta PQR = 12$
C
Area of $\Delta XYZ = 96$,Area of $\Delta PQR = 12$
D
Area of $\Delta XYZ = 48$,Area of $\Delta PQR = 6$
Solution
(A) $1$. By the Midpoint Theorem,the triangle formed by joining the midpoints of the sides of a triangle has an area equal to $1/4$ of the area of the original triangle.
$2$. For $\Delta XYZ$,$\Delta ABC$ is formed by joining the midpoints of its sides. Therefore,$\text{Area}(\Delta ABC) = (1/4) \times \text{Area}(\Delta XYZ)$.
$3$. Given $\text{Area}(\Delta ABC) = 24$,we have $24 = (1/4) \times \text{Area}(\Delta XYZ)$,which implies $\text{Area}(\Delta XYZ) = 24 \times 4 = 96$.
$4$. Similarly,for $\Delta ABC$,$\Delta PQR$ is formed by joining the midpoints of its sides. Therefore,$\text{Area}(\Delta PQR) = (1/4) \times \text{Area}(\Delta ABC)$.
$5$. Substituting the given value,$\text{Area}(\Delta PQR) = (1/4) \times 24 = 6$.
$6$. Thus,the area of $\Delta XYZ$ is $96$ and the area of $\Delta PQR$ is $6$.
$2$. For $\Delta XYZ$,$\Delta ABC$ is formed by joining the midpoints of its sides. Therefore,$\text{Area}(\Delta ABC) = (1/4) \times \text{Area}(\Delta XYZ)$.
$3$. Given $\text{Area}(\Delta ABC) = 24$,we have $24 = (1/4) \times \text{Area}(\Delta XYZ)$,which implies $\text{Area}(\Delta XYZ) = 24 \times 4 = 96$.
$4$. Similarly,for $\Delta ABC$,$\Delta PQR$ is formed by joining the midpoints of its sides. Therefore,$\text{Area}(\Delta PQR) = (1/4) \times \text{Area}(\Delta ABC)$.
$5$. Substituting the given value,$\text{Area}(\Delta PQR) = (1/4) \times 24 = 6$.
$6$. Thus,the area of $\Delta XYZ$ is $96$ and the area of $\Delta PQR$ is $6$.
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View Solution100
MediumMCQ
In $\Delta ABC,$ the midpoints of the sides are $P, Q, R.$ In $\Delta PQR,$ the midpoints of the sides are $X, Y, Z.$ If the area of $\Delta XYZ$ is $20,$ find the area of $\Delta PQR$ and $\Delta ABC.$
A
Area of $\Delta PQR = 80,$ Area of $\Delta ABC = 320$
B
Area of $\Delta PQR = 40,$ Area of $\Delta ABC = 160$
C
Area of $\Delta PQR = 160,$ Area of $\Delta ABC = 640$
D
Area of $\Delta PQR = 80,$ Area of $\Delta ABC = 160$
Solution
(A) The midpoint theorem states that the triangle formed by joining the midpoints of a triangle has an area equal to $1/4$ of the area of the original triangle.
Let $\text{Area}(\Delta PQR) = A_1$ and $\text{Area}(\Delta ABC) = A_2$.
Since $P, Q, R$ are midpoints of $\Delta ABC,$ $\text{Area}(\Delta PQR) = \frac{1}{4} \text{Area}(\Delta ABC).$
Similarly,since $X, Y, Z$ are midpoints of $\Delta PQR,$ $\text{Area}(\Delta XYZ) = \frac{1}{4} \text{Area}(\Delta PQR).$
Given $\text{Area}(\Delta XYZ) = 20.$
Therefore,$20 = \frac{1}{4} \text{Area}(\Delta PQR) \implies \text{Area}(\Delta PQR) = 20 \times 4 = 80.$
Now,$\text{Area}(\Delta PQR) = \frac{1}{4} \text{Area}(\Delta ABC) \implies 80 = \frac{1}{4} \text{Area}(\Delta ABC) \implies \text{Area}(\Delta ABC) = 80 \times 4 = 320.$
Thus,the area of $\Delta PQR$ is $80$ and the area of $\Delta ABC$ is $320.$
Let $\text{Area}(\Delta PQR) = A_1$ and $\text{Area}(\Delta ABC) = A_2$.
Since $P, Q, R$ are midpoints of $\Delta ABC,$ $\text{Area}(\Delta PQR) = \frac{1}{4} \text{Area}(\Delta ABC).$
Similarly,since $X, Y, Z$ are midpoints of $\Delta PQR,$ $\text{Area}(\Delta XYZ) = \frac{1}{4} \text{Area}(\Delta PQR).$
Given $\text{Area}(\Delta XYZ) = 20.$
Therefore,$20 = \frac{1}{4} \text{Area}(\Delta PQR) \implies \text{Area}(\Delta PQR) = 20 \times 4 = 80.$
Now,$\text{Area}(\Delta PQR) = \frac{1}{4} \text{Area}(\Delta ABC) \implies 80 = \frac{1}{4} \text{Area}(\Delta ABC) \implies \text{Area}(\Delta ABC) = 80 \times 4 = 320.$
Thus,the area of $\Delta PQR$ is $80$ and the area of $\Delta ABC$ is $320.$
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