Similarity of Triangles and Basic Proportionality Theorem Questions in English

(N/A) Given: In $\Delta ABC$,$DE || BC$,where $D$ and $E$ are points on $AB$ and $AC$ respectively.
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.
Therefore,$\frac{AD}{DB} = \frac{AE}{EC}$.
Taking the reciprocal of both sides:
$\frac{DB}{AD} = \frac{EC}{AE}$.
Adding $1$ to both sides:
$\frac{DB}{AD} + 1 = \frac{EC}{AE} + 1$.
Taking the common denominator:
$\frac{DB + AD}{AD} = \frac{EC + AE}{AE}$.
Since $DB + AD = AB$ and $EC + AE = AC$,we get:
$\frac{AB}{AD} = \frac{AC}{AE}$.
Taking the reciprocal again:
$\frac{AD}{AB} = \frac{AE}{AC}$.
Hence,it is proved.

(N/A) Let us join $AC$ to intersect $EF$ at $G$ (see Figure).
Given: $AB \parallel DC$ and $EF \parallel AB$.
Since lines parallel to the same line are parallel to each other,we have $EF \parallel DC$.
Now,in $\Delta ADC$,since $EG \parallel DC$ (as $EF \parallel DC$),by Basic Proportionality Theorem,we have:
$\frac{AE}{ED} = \frac{AG}{GC} \quad ...(1)$
Similarly,in $\Delta CAB$,since $GF \parallel AB$ (as $EF \parallel AB$),by Basic Proportionality Theorem,we have:
$\frac{CG}{AG} = \frac{CF}{BF}$
Taking the reciprocal of both sides,we get:
$\frac{AG}{GC} = \frac{BF}{FC} \quad ...(2)$
From equations $(1)$ and $(2)$,we get:
$\frac{AE}{ED} = \frac{BF}{FC}$.

(N/A) It is given that $\frac{PS}{SQ} = \frac{PT}{TR}$.
By the Converse of Thales Theorem (Basic Proportionality Theorem),$ST \parallel QR$.
Therefore,$\angle PST = \angle PQR$ (Corresponding angles) $...(1)$
Also,it is given that $\angle PST = \angle PRQ$ $...(2)$
From equations $(1)$ and $(2)$,we get $\angle PQR = \angle PRQ$.
Since the angles opposite to the sides $PQ$ and $PR$ are equal,the sides themselves must be equal.
Therefore,$PQ = PR$.
Hence,$\triangle PQR$ is an isosceles triangle.

(N/A) $(i)$ Let $EC = x \text{ cm}.$
It is given that $DE || BC.$
By using the Basic Proportionality Theorem,we obtain:
$\frac{AD}{DB} = \frac{AE}{EC}$
$\frac{1.5}{3} = \frac{1}{x}$
$x = \frac{3 \times 1}{1.5}$
$x = 2$
$\therefore EC = 2 \text{ cm}.$
$(ii)$ Let $AD = x \text{ cm}.$
It is given that $DE || BC.$
By using the Basic Proportionality Theorem,we obtain:
$\frac{AD}{DB} = \frac{AE}{EC}$
$\frac{x}{7.2} = \frac{1.8}{5.4}$
$x = \frac{1.8 \times 7.2}{5.4}$
$x = 2.4$
$\therefore AD = 2.4 \text{ cm}.$

(N/A) According to the Converse of Thales Theorem (Basic Proportionality Theorem),$EF || QR$ if and only if $\frac{PE}{EQ} = \frac{PF}{FR}$.
Given values are:
$PE = 3.9 \ cm$
$EQ = 3 \ cm$
$PF = 3.6 \ cm$
$FR = 2.4 \ cm$
Calculating the ratios:
$\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$
$\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5$
Since $\frac{PE}{EQ} \neq \frac{PF}{FR}$ $(1.3 \neq 1.5)$,the condition for $EF$ being parallel to $QR$ is not satisfied.
Therefore,$EF$ is not parallel to $QR$.

(N/A) Given:
$PE = 4 \, cm, QE = 4.5 \, cm, PF = 8 \, cm, RF = 9 \, cm$.
According to the Converse of the Basic Proportionality Theorem (Thales Theorem),if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.
Calculate the ratios:
$\frac{PE}{EQ} = \frac{4}{4.5} = \frac{40}{45} = \frac{8}{9}$.
$\frac{PF}{FR} = \frac{8}{9}$.
Since $\frac{PE}{EQ} = \frac{PF}{FR}$,by the Converse of the Basic Proportionality Theorem,$EF$ is parallel to $QR$ $(EF || QR)$.

(A) Given: $PQ = 1.28 \, cm, PR = 2.56 \, cm, PE = 0.18 \, cm, PF = 0.36 \, cm$.
According to 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.
Calculate the ratios:
$\frac{PE}{PQ} = \frac{0.18}{1.28} = \frac{18}{128} = \frac{9}{64}$
$\frac{PF}{PR} = \frac{0.36}{2.56} = \frac{36}{256} = \frac{9}{64}$
Since $\frac{PE}{PQ} = \frac{PF}{PR} = \frac{9}{64}$,the line $EF$ divides the sides $PQ$ and $PR$ in the same ratio.
Therefore,by the converse of the Basic Proportionality Theorem,$EF || QR$.

(N/A) In the given figure,$LM \parallel CB$.
By using the Basic Proportionality Theorem (Thales Theorem) in $\triangle ABC$,we obtain:
$\frac{AM}{AB} = \frac{AL}{AC} \quad ...(i)$
Similarly,in $\triangle ADC$,since $LN \parallel CD$,by using the Basic Proportionality Theorem,we obtain:
$\frac{AN}{AD} = \frac{AL}{AC} \quad ...(ii)$
From equation $(i)$ and equation $(ii)$,we observe that the right-hand sides are equal.
Therefore,we obtain:
$\frac{AM}{AB} = \frac{AN}{AD}$

(N/A) In $\Delta ABC$,$DE \parallel AC$.
By the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{BD}{DA} = \frac{BE}{EC} \quad ...(i)$
In $\Delta BAE$,$DF \parallel AE$.
By the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{BD}{DA} = \frac{BF}{FE} \quad ...(ii)$
From equations $(i)$ and $(ii)$,since the left-hand sides are equal,the right-hand sides must also be equal:
$\frac{BF}{FE} = \frac{BE}{EC}$
Hence,it is proved.

(N/A) In $\Delta POQ$,$DE \parallel OQ$.
Therefore,$\frac{PE}{EQ} = \frac{PD}{DO}$ (Basic Proportionality Theorem) $...(i)$
In $\Delta POR$,$DF \parallel OR$.
Therefore,$\frac{PF}{FR} = \frac{PD}{DO}$ (Basic Proportionality Theorem) $...(ii)$
From $(i)$ and $(ii)$,we obtain:
$\frac{PE}{EQ} = \frac{PF}{FR}$
Therefore,$EF \parallel QR$ (Converse of Basic Proportionality Theorem).

(N/A) In $\Delta POQ$,$AB \parallel PQ$.
Therefore,$\frac{OA}{AP} = \frac{OB}{BQ}$ (Basic Proportionality Theorem) $...(i)$
In $\Delta POR$,$AC \parallel PR$.
Therefore,$\frac{OA}{AP} = \frac{OC}{CR}$ (Basic Proportionality Theorem) $...(ii)$
From $(i)$ and $(ii)$,we obtain:
$\frac{OB}{BQ} = \frac{OC}{CR}$
Therefore,$BC \parallel QR$ (By the converse of the Basic Proportionality Theorem).

(N/A) Consider a triangle $ABC$ in which $P$ is the mid-point of side $AB$ and a line $PQ$ is drawn parallel to $BC$ such that it intersects $AC$ at $Q$.
According to the Basic Proportionality Theorem (Theorem $6.1$),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.
Therefore,we have:
$\frac{AP}{PB} = \frac{AQ}{QC}$
Since $P$ is the mid-point of $AB$,we have $AP = PB$,which implies $\frac{AP}{PB} = 1$.
Substituting this into the equation,we get:
$1 = \frac{AQ}{QC}$
$\Rightarrow AQ = QC$
This proves that $Q$ is the mid-point of $AC$,meaning the line $PQ$ bisects the third side $AC$.

(N/A) Consider a triangle $ABC$ in which $P$ and $Q$ are the mid-points of sides $AB$ and $AC$ respectively.
Given: $AP = PB$ and $AQ = QC$.
To prove: $PQ \parallel BC$.
Proof:
Since $P$ is the mid-point of $AB$,we have $AP = PB$,which implies $\frac{AP}{PB} = 1$.
Since $Q$ is the mid-point of $AC$,we have $AQ = QC$,which implies $\frac{AQ}{QC} = 1$.
From the above two equations,we get $\frac{AP}{PB} = \frac{AQ}{QC}$.
According to the converse of the Basic Proportionality Theorem (Theorem $6.2$),if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.
Therefore,$PQ \parallel BC$.

(N/A) Draw a line $EF$ through point $O$,such that $EF \parallel CD$.
In $\triangle ADC$,since $EO \parallel CD$,by using the Basic Proportionality Theorem $(BPT)$,we obtain:
$\frac{AE}{ED} = \frac{AO}{OC} \quad ...(1)$
Since $AB \parallel CD$ and $EF \parallel CD$,it follows that $EF \parallel AB$.
In $\triangle ABD$,since $EO \parallel AB$,by using the Basic Proportionality Theorem,we obtain:
$\frac{AE}{ED} = \frac{BO}{OD} \quad ...(2)$
From equations $(1)$ and $(2)$,we have:
$\frac{AO}{OC} = \frac{BO}{OD}$
Rearranging the terms,we get:
$\frac{AO}{BO} = \frac{OC}{OD}$

(N/A) Let us consider the following figure for the given question.
Draw a line $OE \parallel AB$ such that $E$ lies on $AD$.
In $\triangle ABD$,since $OE \parallel AB$,by using the Basic Proportionality Theorem $(BPT)$,we obtain:
$\frac{AE}{ED} = \frac{BO}{OD}$ $...(1)$
However,it is given that:
$\frac{AO}{BO} = \frac{CO}{DO}$
Rearranging the terms,we get:
$\frac{AO}{CO} = \frac{BO}{DO}$ $...(2)$
From equations $(1)$ and $(2)$,we obtain:
$\frac{AE}{ED} = \frac{AO}{CO}$
In $\triangle ADC$,since $\frac{AE}{ED} = \frac{AO}{OC}$,by the converse of the Basic Proportionality Theorem,we have:
$EO \parallel DC$
Since we constructed $OE \parallel AB$ and we proved $OE \parallel DC$,it follows that:
$AB \parallel DC$
Therefore,$ABCD$ is a trapezium because one pair of opposite sides is parallel.

(N/A) Let $AB$ denote the lamp-post and $CD$ the girl after walking for $4\, seconds$ away from the lamp-post.
From the figure,you can see that $DE$ is the shadow of the girl. Let $DE$ be $x$ meters.
Now,$BD = 1.2\, m/s \times 4\, s = 4.8\, m$.
Note that in $\Delta ABE$ and $\Delta CDE$:
$\angle B = \angle D = 90^{\circ}$ (Each is $90^{\circ}$ because the lamp-post and the girl are standing vertical to the ground).
$\angle E = \angle E$ (Common angle).
So,$\Delta ABE \sim \Delta CDE$ ($AA$ similarity criterion).
Therefore,$\frac{BE}{DE} = \frac{AB}{CD}$.
Given $AB = 3.6\, m$ and $CD = 90\, cm = 0.9\, m$.
$\frac{4.8 + x}{x} = \frac{3.6}{0.9}$.
$\frac{4.8 + x}{x} = 4$.
$4.8 + x = 4x$.
$3x = 4.8$.
$x = 1.6\, m$.
So,the length of the shadow of the girl after walking for $4\, seconds$ is $1.6\, m$.

(B) Given that $DE \parallel BC$ in $\triangle ABC$,by the Basic Proportionality Theorem (Thales Theorem),$\triangle ADE \sim \triangle ABC$.
Therefore,the ratio of corresponding sides is equal: $\frac{AD}{AB} = \frac{DE}{BC}$.
Given $AD = 2 \, cm$ and $BD = 3 \, cm$,we have $AB = AD + BD = 2 + 3 = 5 \, cm$.
Substituting the values into the ratio: $\frac{2}{5} = \frac{DE}{7.5}$.
Solving for $DE$: $DE = \frac{2}{5} \times 7.5$.
$DE = 2 \times 1.5 = 3 \, cm$.

(A) To determine if $\triangle ABC$ is a right triangle,we check if it satisfies the converse of the Pythagoras theorem,which states that if the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides,then the triangle is a right triangle.
Given sides are $AB = 24 \, cm$,$BC = 10 \, cm$,and $AC = 26 \, cm$.
Calculate the squares of the sides:
$AB^2 = 24^2 = 576 \, cm^2$
$BC^2 = 10^2 = 100 \, cm^2$
$AC^2 = 26^2 = 676 \, cm^2$
Now,check the sum of the squares of the two smaller sides:
$AB^2 + BC^2 = 576 + 100 = 676 \, cm^2$
Since $AB^2 + BC^2 = AC^2$ $(576 + 100 = 676)$,the triangle satisfies the Pythagoras theorem.
Therefore,$\triangle ABC$ is a right triangle with the right angle at vertex $B$.

(B) According to the Converse of the Basic Proportionality Theorem (Thales Theorem),$PQ \parallel EF$ if and only if $\frac{DP}{PE} = \frac{DQ}{QF}$.
Given $DP = 5 \, cm$ and $DE = 15 \, cm$,we find $PE = DE - DP = 15 - 5 = 10 \, cm$.
Thus,the ratio $\frac{DP}{PE} = \frac{5}{10} = \frac{1}{2}$.
Given $DQ = 6 \, cm$ and $QF = 18 \, cm$,the ratio $\frac{DQ}{QF} = \frac{6}{18} = \frac{1}{3}$.
Since $\frac{DP}{PE} \neq \frac{DQ}{QF}$ (i.e.,$\frac{1}{2} \neq \frac{1}{3}$),the line $PQ$ is not parallel to $EF$.

(A) Given: $PQ = 12.5 \, cm$,$PA = 5 \, cm$,$BR = 6 \, cm$,and $PB = 4 \, cm$.
First,calculate the length of $AQ$:
$AQ = PQ - PA = 12.5 - 5 = 7.5 \, cm$.
Now,calculate the ratios:
$\frac{PA}{AQ} = \frac{5}{7.5} = \frac{50}{75} = \frac{2}{3}$ ......$(i)$
$\frac{PB}{BR} = \frac{4}{6} = \frac{2}{3}$ ......$(ii)$
From equations $(i)$ and $(ii)$,we observe that:
$\frac{PA}{AQ} = \frac{PB}{BR}$
According to the converse of the Basic Proportionality Theorem (Thales Theorem),if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.
Therefore,$AB \parallel QR$.

(N/A) Given: $\frac{AD}{DB} = \frac{AE}{EC}$ and $\angle ADE = \angle AED$.
According to the converse of the Basic Proportionality Theorem (Thales Theorem),if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.
Therefore,$DE \parallel BC$.
Since $DE \parallel BC$ and $AB$ is a transversal,the corresponding angles are equal:
$\angle ADE = \angle ABC$ ....... $(1)$
Similarly,since $DE \parallel BC$ and $AC$ is a transversal:
$\angle AED = \angle ACB$ ....... $(2)$
We are given that $\angle ADE = \angle AED$.
Substituting the values from $(1)$ and $(2)$ into this equation,we get:
$\angle ABC = \angle ACB$.
In $\triangle ABC$,since the base angles are equal $(\angle B = \angle C)$,the sides opposite to these angles must also be equal.
Therefore,$AB = AC$.
Since two sides of $\triangle ABC$ are equal,$\triangle ABC$ is an isosceles triangle.

(B) Given,$DE \parallel AB$.
By 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.
Therefore,$\frac{CD}{AD} = \frac{CE}{BE}$.
Substituting the given values:
$\frac{x+3}{3x+19} = \frac{x}{3x+4}$
Cross-multiplying:
$(x+3)(3x+4) = x(3x+19)$
$3x^2 + 4x + 9x + 12 = 3x^2 + 19x$
$3x^2 + 13x + 12 = 3x^2 + 19x$
Subtracting $3x^2$ from both sides:
$13x + 12 = 19x$
$19x - 13x = 12$
$6x = 12$
$x = \frac{12}{6} = 2$.
Hence,the required value of $x$ is $2$.

(N/A) Given: In $\Delta ABC$,$AD \perp BC$.
To prove: $AB^2 + CD^2 = BD^2 + AC^2$.
Proof:
In right-angled $\Delta ADB$,by Pythagoras theorem:
$AB^2 = AD^2 + BD^2$ --- $(1)$
In right-angled $\Delta ADC$,by Pythagoras theorem:
$AC^2 = AD^2 + CD^2$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$AB^2 - AC^2 = (AD^2 + BD^2) - (AD^2 + CD^2)$
$AB^2 - AC^2 = AD^2 + BD^2 - AD^2 - CD^2$
$AB^2 - AC^2 = BD^2 - CD^2$
Rearranging the terms,we get:
$AB^2 + CD^2 = BD^2 + AC^2$
Hence proved.

(N/A) $BL$ and $CM$ are medians of $\Delta ABC$ in which $\angle A = 90^{\circ}$.
From $\Delta ABC$,by Pythagoras theorem:
$BC^2 = AB^2 + AC^2$ $...(1)$
From $\Delta ABL$,since $L$ is the midpoint of $AC$,$AL = AC/2$:
$BL^2 = AB^2 + AL^2 = AB^2 + (AC/2)^2 = AB^2 + AC^2/4$
$4BL^2 = 4AB^2 + AC^2$ $...(2)$
From $\Delta CMA$,since $M$ is the midpoint of $AB$,$AM = AB/2$:
$CM^2 = AC^2 + AM^2 = AC^2 + (AB/2)^2 = AC^2 + AB^2/4$
$4CM^2 = 4AC^2 + AB^2$ $...(3)$
Adding $(2)$ and $(3)$:
$4(BL^2 + CM^2) = 4AB^2 + AC^2 + 4AC^2 + AB^2$
$4(BL^2 + CM^2) = 5AB^2 + 5AC^2$
$4(BL^2 + CM^2) = 5(AB^2 + AC^2)$
Using equation $(1)$,$AB^2 + AC^2 = BC^2$,so:
$4(BL^2 + CM^2) = 5BC^2$.

(N/A) Through $O$,draw $PQ \parallel BC$ such that $P$ lies on $AB$ and $Q$ lies on $DC$.
Now,$PQ \parallel BC$.
Therefore,$PQ \perp AB$ and $PQ \perp DC$ (since $\angle B = 90^{\circ}$ and $\angle C = 90^{\circ}$).
So,$\angle BPQ = 90^{\circ}$ and $\angle CQP = 90^{\circ}$.
Therefore,$BPQC$ and $APQD$ are both rectangles.
Now,from $\Delta OPB$,by Pythagoras theorem:
$OB^{2} = BP^{2} + OP^{2}$ $...(1)$
Similarly,from $\Delta OQD$:
$OD^{2} = OQ^{2} + DQ^{2}$ $...(2)$
From $\Delta OQC$,we have:
$OC^{2} = OQ^{2} + CQ^{2}$ $...(3)$
And from $\Delta OAP$,we have:
$OA^{2} = AP^{2} + OP^{2}$ $...(4)$
Adding $(1)$ and $(2)$:
$OB^{2} + OD^{2} = BP^{2} + OP^{2} + OQ^{2} + DQ^{2}$
Since $BP = CQ$ and $DQ = AP$ (opposite sides of rectangles $BPQC$ and $APQD$):
$OB^{2} + OD^{2} = CQ^{2} + OP^{2} + OQ^{2} + AP^{2}$
Rearranging the terms:
$OB^{2} + OD^{2} = (CQ^{2} + OQ^{2}) + (AP^{2} + OP^{2})$
Using equations $(3)$ and $(4)$:
$OB^{2} + OD^{2} = OC^{2} + OA^{2}$.

(A) The given sides of the triangle are $7 \, cm, 24 \, cm,$ and $25 \, cm$.
To check if it is a right-angled triangle,we use the converse of the Pythagoras theorem,which states that if the square of the longest side is equal to the sum of the squares of the other two sides,then the triangle is a right-angled triangle.
First,we calculate the squares of the given sides:
$7^2 = 49$
$24^2 = 576$
$25^2 = 625$
Now,we check if the sum of the squares of the two smaller sides equals the square of the largest side:
$49 + 576 = 625$
Since $7^2 + 24^2 = 25^2$,the given sides satisfy the Pythagoras theorem.
Therefore,the triangle is a right-angled triangle.
The longest side of a right-angled triangle is its hypotenuse. Thus,the length of the hypotenuse is $25 \, cm$.

(N/A) The given sides of the triangle are $3 \text{ cm}, 8 \text{ cm},$ and $6 \text{ cm}$.
According to the converse of the Pythagoras theorem,a triangle is a right triangle if the sum of the squares of the two smaller sides is equal to the square of the longest side.
Here,the squares of the sides are:
$3^2 = 9$
$6^2 = 36$
$8^2 = 64$
Checking the sum of the squares of the two smaller sides:
$3^2 + 6^2 = 9 + 36 = 45$
Since $45 \neq 64$ (i.e.,$3^2 + 6^2 \neq 8^2$),the condition for a right triangle is not satisfied.
Therefore,the given triangle is not a right triangle.

(NONE) Given that the sides of the triangle are $50 \text{ cm}, 80 \text{ cm},$ and $100 \text{ cm}.$
To check if it is a right-angled triangle,we apply the converse of the Pythagoras theorem,which states that in a triangle,if the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides,then it is a right-angled triangle.
Calculate the squares of the sides:
$50^2 = 2500$
$80^2 = 6400$
$100^2 = 10000$
Now,check the sum of the squares of the two smaller sides:
$50^2 + 80^2 = 2500 + 6400 = 8900$
Comparing this with the square of the longest side:
$8900 \neq 10000$
Since $50^2 + 80^2 \neq 100^2$,the given triangle does not satisfy the Pythagoras theorem.
Therefore,the given triangle is not a right-angled triangle.

(A) Given that the sides are $13\, cm, 12\, cm,$ and $5\, cm$.
Squaring the lengths of these sides,we obtain $169, 144,$ and $25$.
Clearly,$144 + 25 = 169$.
Or,$12^2 + 5^2 = 13^2$.
The sides of the given triangle satisfy the Pythagoras theorem.
Therefore,it is a right triangle.
We know that the longest side of a right triangle is the hypotenuse.
Therefore,the length of the hypotenuse of this triangle is $13\, cm$.

(N/A) Given that $\triangle ABC$ is an isosceles triangle with $\angle C = 90^{\circ}$.
Since it is an isosceles triangle,the two legs must be equal,so $AC = CB$.
Applying the Pythagoras theorem in $\triangle ABC$ (which is right-angled at point $C$),we have:
$AC^{2} + CB^{2} = AB^{2}$
Since $AC = CB$,we can substitute $CB$ with $AC$ in the equation:
$AC^{2} + AC^{2} = AB^{2}$
$2AC^{2} = AB^{2}$
Hence,it is proved that $AB^{2} = 2AC^{2}$.

(N/A) Given that,
$AB^2 = 2 AC^2$
We can rewrite this as:
$AB^2 = AC^2 + AC^2$
Since it is given that $AC = BC$,we can substitute $BC$ for one of the $AC$ terms:
$AB^2 = AC^2 + BC^2$
This equation satisfies the converse of the Pythagoras theorem,which states that if the square of one side of a triangle is equal to the sum of the squares of the other two sides,then the angle opposite the first side is a right angle.
Therefore,$\triangle ABC$ is a right-angled triangle with the right angle at $C$.

(N/A) Let $AD$ be the altitude in the given equilateral triangle,$\triangle ABC$.
We know that an altitude in an equilateral triangle bisects the opposite side.
$\therefore BD = DC = a$
In $\triangle ADB$,we have $\angle ADB = 90^{\circ}$.
Applying the Pythagoras theorem,we obtain:
$AD^2 + DB^2 = AB^2$
$\Rightarrow AD^2 + a^2 = (2a)^2$
$\Rightarrow AD^2 + a^2 = 4a^2$
$\Rightarrow AD^2 = 3a^2$
$\Rightarrow AD = a\sqrt{3}$
In an equilateral triangle,all the altitudes are equal in length.
Therefore,the length of each altitude is $a\sqrt{3}$.

(N/A) Let $ABCD$ be a rhombus with diagonals $AC$ and $BD$ intersecting at $O$.
In $\Delta AOB, \Delta BOC, \Delta COD,$ and $\Delta AOD,$
Applying the Pythagoras theorem,we obtain:
$AB^2 = AO^2 + OB^2$ $...(1)$
$BC^2 = BO^2 + OC^2$ $...(2)$
$CD^2 = CO^2 + OD^2$ $...(3)$
$AD^2 = AO^2 + OD^2$ $...(4)$
Adding all these equations,we obtain:
$AB^2 + BC^2 + CD^2 + AD^2 = 2(AO^2 + OB^2 + OC^2 + OD^2)$
Since the diagonals of a rhombus bisect each other at right angles,$AO = OC = AC/2$ and $BO = OD = BD/2$.
Substituting these values:
$= 2((AC/2)^2 + (BD/2)^2 + (AC/2)^2 + (BD/2)^2)$
$= 2(2(AC/2)^2 + 2(BD/2)^2)$
$= 2(AC^2/2 + BD^2/2)$
$= AC^2 + BD^2$
Thus,the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

(A) Let $OA$ be the wall and $AB$ be the ladder,where $OA = 8\, m$ and $AB = 10\, m$.
Since the wall is perpendicular to the ground,$\triangle AOB$ is a right-angled triangle with $\angle AOB = 90^\circ$.
By Pythagoras theorem,$AB^2 = OA^2 + OB^2$.
Substituting the values: $(10)^2 = (8)^2 + OB^2$.
$100 = 64 + OB^2$.
$OB^2 = 100 - 64 = 36$.
$OB = \sqrt{36} = 6\, m$.
Therefore,the distance of the foot of the ladder from the base of the wall is $6\, m$.

(N/A) Let $OB$ be the pole and $AB$ be the wire. The pole is vertical,so $\triangle AOB$ is a right-angled triangle with $\angle AOB = 90^{\circ}$.
By Pythagoras theorem,
$AB^{2} = OB^{2} + OA^{2}$
$(24\, m)^{2} = (18\, m)^{2} + OA^{2}$
$576\, m^{2} = 324\, m^{2} + OA^{2}$
$OA^{2} = (576 - 324)\, m^{2} = 252\, m^{2}$
$OA = \sqrt{252}\, m = \sqrt{36 \times 7}\, m = 6\sqrt{7}\, m$
Therefore,the distance from the base is $6\sqrt{7}\, m$ (approximately $15.87\, m$).

(N/A) Applying the Pythagoras theorem in $\triangle ACE$,we obtain:
$AC^{2} + CE^{2} = AE^{2}$ $...(1)$
Applying the Pythagoras theorem in $\triangle BCD$,we obtain:
$BC^{2} + CD^{2} = BD^{2}$ $...(2)$
Adding equation $(1)$ and equation $(2)$,we obtain:
$AC^{2} + CE^{2} + BC^{2} + CD^{2} = AE^{2} + BD^{2}$ $...(3)$
Applying the Pythagoras theorem in $\triangle CDE$,we obtain:
$DE^{2} = CD^{2} + CE^{2}$ $...(4)$
Applying the Pythagoras theorem in $\triangle ABC$,we obtain:
$AB^{2} = AC^{2} + BC^{2}$ $...(5)$
Substituting the values from equation $(4)$ and equation $(5)$ into equation $(3)$,we obtain:
$AB^{2} + DE^{2} = AE^{2} + BD^{2}$
Hence,$AE^{2} + BD^{2} = AB^{2} + DE^{2}$.

(N/A) Applying the Pythagoras theorem for $\triangle ACD$,we obtain:
$AC^{2} = AD^{2} + DC^{2}$
$AD^{2} = AC^{2} - DC^{2}$ $...(1)$
Applying the Pythagoras theorem for $\triangle ABD$,we obtain:
$AB^{2} = AD^{2} + DB^{2}$
$AD^{2} = AB^{2} - DB^{2}$ $...(2)$
From equation $(1)$ and equation $(2)$,we obtain:
$AC^{2} - DC^{2} = AB^{2} - DB^{2}$ $...(3)$
It is given that $DB = 3CD$. Since $BC = DB + CD$,we have $BC = 3CD + CD = 4CD$.
$\therefore CD = \frac{BC}{4}$ and $DB = 3CD = \frac{3BC}{4}$.
Substituting these values into equation $(3)$,we obtain:
$AC^{2} - \left(\frac{BC}{4}\right)^{2} = AB^{2} - \left(\frac{3BC}{4}\right)^{2}$
$AC^{2} - \frac{BC^{2}}{16} = AB^{2} - \frac{9BC^{2}}{16}$
$16AC^{2} - BC^{2} = 16AB^{2} - 9BC^{2}$
$16AB^{2} - 16AC^{2} = 8BC^{2}$
Dividing by $8$,we get:
$2AB^{2} = 2AC^{2} + BC^{2}$

(N/A) Let the side of the equilateral triangle be $a$,and $AE$ be the altitude of $\Delta ABC$ drawn to $BC$.
Since $AE$ is the altitude of an equilateral triangle,it bisects the base $BC$.
$\therefore BE = EC = \frac{BC}{2} = \frac{a}{2}$.
In the right-angled triangle $\Delta ABE$,by Pythagoras theorem:
$AE^{2} = AB^{2} - BE^{2} = a^{2} - (\frac{a}{2})^{2} = a^{2} - \frac{a^{2}}{4} = \frac{3a^{2}}{4}$.
So,$AE = \frac{a\sqrt{3}}{2}$.
Given that $BD = \frac{1}{3} BC = \frac{a}{3}$.
Now,$DE = BE - BD = \frac{a}{2} - \frac{a}{3} = \frac{3a - 2a}{6} = \frac{a}{6}$.
In the right-angled triangle $\Delta ADE$,by Pythagoras theorem:
$AD^{2} = AE^{2} + DE^{2}$.
Substituting the values:
$AD^{2} = (\frac{a\sqrt{3}}{2})^{2} + (\frac{a}{6})^{2} = \frac{3a^{2}}{4} + \frac{a^{2}}{36}$.
Taking $LCM$ as $36$:
$AD^{2} = \frac{27a^{2} + a^{2}}{36} = \frac{28a^{2}}{36} = \frac{7a^{2}}{9}$.
Since $a = AB$,we have $AD^{2} = \frac{7 AB^{2}}{9}$.
Therefore,$9 AD^{2} = 7 AB^{2}$.
Hence proved.

(N/A) Let the side of the equilateral triangle be $a$,and $AE$ be the altitude of $\triangle ABC$.
Since the altitude of an equilateral triangle bisects the base,
$BE = EC = \frac{BC}{2} = \frac{a}{2}$
Applying the Pythagoras theorem in the right-angled $\triangle ABE$,we obtain:
$AB^2 = AE^2 + BE^2$
Substituting the values $AB = a$ and $BE = \frac{a}{2}$:
$a^2 = AE^2 + (\frac{a}{2})^2$
$a^2 = AE^2 + \frac{a^2}{4}$
Rearranging the terms to solve for $AE^2$:
$AE^2 = a^2 - \frac{a^2}{4}$
$AE^2 = \frac{4a^2 - a^2}{4}$
$AE^2 = \frac{3a^2}{4}$
Multiplying both sides by $4$:
$4AE^2 = 3a^2$
Thus,it is proved that $4 \times$ (square of altitude) $= 3 \times$ (square of one side).

(B) Given that,$AB = 6\sqrt{3} \text{ cm}$,$AC = 12 \text{ cm}$,and $BC = 6 \text{ cm}$.
Calculating the squares of the sides:
$AB^2 = (6\sqrt{3})^2 = 36 \times 3 = 108$
$BC^2 = (6)^2 = 36$
$AC^2 = (12)^2 = 144$
Now,observe that:
$AB^2 + BC^2 = 108 + 36 = 144$
Since $AB^2 + BC^2 = AC^2$,the triangle satisfies the converse of the Pythagoras theorem.
Therefore,$\triangle ABC$ is a right-angled triangle,with the right angle opposite to the hypotenuse $AC$.
Thus,$\angle B = 90^\circ$.
Hence,the correct answer is $(B)$.

(N/A) Construction: Draw a line segment $RT$ parallel to $PS$ which intersects the extended line segment $QP$ at point $T$.
Given that,$PS$ is the angle bisector of $\angle QPR$.
Therefore,$\angle QPS = \angle SPR \quad \dots(1)$
By construction,$PS \parallel TR$ and $QT$ is a transversal.
Therefore,$\angle QPS = \angle QTR$ (Corresponding angles) $\quad \dots(2)$
Also,$PS \parallel TR$ and $PR$ is a transversal.
Therefore,$\angle SPR = \angle PRT$ (Alternate interior angles) $\quad \dots(3)$
From equations $(1)$,$(2)$,and $(3)$,we get:
$\angle QTR = \angle PRT$
In $\Delta PTR$,since $\angle QTR = \angle PRT$,the sides opposite to these angles are equal.
Therefore,$PT = PR \quad \dots(4)$
In $\Delta QTR$,since $PS \parallel TR$,by the Basic Proportionality Theorem $(BPT)$:
$\frac{QS}{SR} = \frac{QP}{PT}$
Substituting $PT = PR$ from equation $(4)$ into the above expression:
$\frac{QS}{SR} = \frac{PQ}{PR}$
Hence,it is proved.

(N/A) In $\triangle ADB$,applying the Pythagoras theorem,we have:
$AB^{2} = AD^{2} + BD^{2} \quad \dots(1)$
In $\triangle ADC$,applying the Pythagoras theorem,we have:
$AC^{2} = AD^{2} + DC^{2}$
Since $DC = DB + BC$,we can write:
$AC^{2} = AD^{2} + (DB + BC)^{2}$
$AC^{2} = AD^{2} + DB^{2} + BC^{2} + 2DB \cdot BC$
Substituting $AD^{2} + DB^{2} = AB^{2}$ from equation $(1)$:
$AC^{2} = AB^{2} + BC^{2} + 2BC \cdot BD$

(N/A) Applying the Pythagoras theorem in $\triangle ADB$,we obtain:
$AD^{2} + DB^{2} = AB^{2}$
$\Rightarrow AD^{2} = AB^{2} - DB^{2} \dots(1)$
Applying the Pythagoras theorem in $\triangle ADC$,we obtain:
$AD^{2} + DC^{2} = AC^{2}$
Substituting the value of $AD^{2}$ from equation $(1)$:
$(AB^{2} - BD^{2}) + DC^{2} = AC^{2}$
Since $D$ lies on $BC$,we have $DC = BC - BD$. Substituting this into the equation:
$AB^{2} - BD^{2} + (BC - BD)^{2} = AC^{2}$
$AB^{2} - BD^{2} + (BC^{2} + BD^{2} - 2BC \cdot BD) = AC^{2}$
$AB^{2} + BC^{2} - 2BC \cdot BD = AC^{2}$
Thus,$AC^{2} = AB^{2} + BC^{2} - 2BC \cdot BD$.

(N/A) Applying Pythagoras theorem in $\Delta AMD$,we obtain:
$AM^{2} + MD^{2} = AD^{2} \dots(1)$
Applying Pythagoras theorem in $\Delta AMC$,we obtain:
$AM^{2} + MC^{2} = AC^{2}$
Since $MC = MD + DC$,we have:
$AM^{2} + (MD + DC)^{2} = AC^{2}$
$AM^{2} + MD^{2} + DC^{2} + 2MD \cdot DC = AC^{2}$
Substituting $AM^{2} + MD^{2} = AD^{2}$ from equation $(1)$:
$AD^{2} + DC^{2} + 2MD \cdot DC = AC^{2}$
Since $AD$ is a median,$DC = \frac{BC}{2}$. Substituting this value:
$AD^{2} + \left(\frac{BC}{2}\right)^{2} + 2MD \cdot \left(\frac{BC}{2}\right) = AC^{2}$
$AD^{2} + \left(\frac{BC}{2}\right)^{2} + BC \cdot DM = AC^{2}$
Thus,$AC^{2} = AD^{2} + BC \cdot DM + \left(\frac{BC}{2}\right)^{2}$.

(A) Let $ABCD$ be a parallelogram.
Let us draw perpendicular $DE$ on the extended side $AB$,and $AF$ on side $DC$.
Applying Pythagoras theorem in $\triangle DEA$,we obtain:
$DE^{2} + EA^{2} = DA^{2} \dots (i)$
Applying Pythagoras theorem in $\triangle DEB$,we obtain:
$DE^{2} + EB^{2} = DB^{2}$
$DE^{2} + (EA + AB)^{2} = DB^{2}$
$(DE^{2} + EA^{2}) + AB^{2} + 2EA \times AB = DB^{2}$
$DA^{2} + AB^{2} + 2EA \times AB = DB^{2} \dots (ii)$
Applying Pythagoras theorem in $\triangle AFC$,we obtain:
$AC^{2} = AF^{2} + FC^{2}$
$AC^{2} = AF^{2} + (DC - FD)^{2}$
$AC^{2} = AF^{2} + DC^{2} + FD^{2} - 2DC \times FD$
$AC^{2} = (AF^{2} + FD^{2}) + DC^{2} - 2DC \times FD$
$AC^{2} = AD^{2} + DC^{2} - 2DC \times FD \dots (iii)$
Since $ABCD$ is a parallelogram,$AB = CD \dots (iv)$ and $BC = AD \dots (v)$.
In $\triangle DEA$ and $\triangle AFD$:
$\angle DEA = \angle AFD = 90^{\circ}$
$\angle EAD = \angle ADF$ (since $EA \parallel DF$)
$AD = AD$ (Common side)
Therefore,$\triangle EAD \cong \triangle FDA$ ($AAS$ congruence criterion).
This implies $EA = DF \dots (vi)$.
Adding equations $(ii)$ and $(iii)$:
$DB^{2} + AC^{2} = DA^{2} + AB^{2} + 2EA \times AB + AD^{2} + DC^{2} - 2DC \times FD$
Using $AB = DC$ and $EA = DF$:
$DB^{2} + AC^{2} = AD^{2} + AB^{2} + 2EA \times AB + AD^{2} + AB^{2} - 2AB \times EA$
$DB^{2} + AC^{2} = 2AD^{2} + 2AB^{2}$
Since $AD = BC$ and $AB = CD$:
$AC^{2} + BD^{2} = AB^{2} + BC^{2} + CD^{2} + DA^{2}$.

(N/A) Let us extend $BA$ to $P$ such that $AP = AC$. Join $PC$.
It is given that,
$\frac{BD}{CD} = \frac{AB}{AC}$
Since $AP = AC$,we can substitute $AP$ for $AC$:
$\frac{BD}{CD} = \frac{AB}{AP}$
By using the converse of the Basic Proportionality Theorem $(BPT)$ in $\Delta BPC$,we obtain:
$AD \parallel PC$
Therefore,$\angle BAD = \angle APC$ (Corresponding angles) $\dots(1)$
And,$\angle DAC = \angle ACP$ (Alternate interior angles) $\dots(2)$
By construction,we have $AP = AC$,which implies that in $\Delta APC$,the angles opposite to equal sides are equal:
$\angle APC = \angle ACP \dots(3)$
From equations $(1), (2),$ and $(3)$,we get:
$\angle BAD = \angle DAC$
Thus,$AD$ is the bisector of $\angle BAC$.

(1:16) Given that,$\frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{4}$.
Since $\frac{AD}{AB} = \frac{AE}{AC}$,by the Converse of Thales Theorem (Basic Proportionality Theorem),$DE \parallel BC$.
Therefore,$\Delta ADE \sim \Delta ABC$ by $AA$ similarity criterion.
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Thus,$\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta ABC)} = \left(\frac{AD}{AB}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.
Now,calculate the area of $\Delta ABC$ using the coordinates $A(4, 6)$,$B(1, 5)$,and $C(7, 2)$:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Area $= \frac{1}{2} |4(5 - 2) + 1(2 - 6) + 7(6 - 5)|$
Area $= \frac{1}{2} |4(3) + 1(-4) + 7(1)| = \frac{1}{2} |12 - 4 + 7| = \frac{1}{2} |15| = 7.5$ square units.
Area of $\Delta ADE = \frac{1}{16} \times \text{Area}(\Delta ABC) = \frac{1}{16} \times 7.5 = \frac{7.5}{16} = \frac{15}{32}$ square units.
The ratio of the area of $\Delta ADE$ to the area of $\Delta ABC$ is $1:16$.

(A) Given,a trapezium $ABCD$ in which $AB \parallel DC$. $P$ and $Q$ are points on $AD$ and $BC$ respectively such that $PQ \parallel DC$. Thus,$AB \parallel PQ \parallel DC$.
Join $BD$. Let $BD$ intersect $PQ$ at point $O$.
In $\triangle ABD$,$PO \parallel AB$ (since $PQ \parallel AB$).
By the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{AP}{PD} = \frac{BO}{OD}$ .......$(i)$
In $\triangle BDC$,$OQ \parallel DC$ (since $PQ \parallel DC$).
By the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{BQ}{QC} = \frac{BO}{OD}$ .......$(ii)$
From equations $(i)$ and $(ii)$,we get:
$\frac{AP}{PD} = \frac{BQ}{QC}$
Substitute the given values $PD = 18 \, cm$,$BQ = 35 \, cm$,and $QC = 15 \, cm$:
$\frac{AP}{18} = \frac{35}{15}$
$AP = \frac{35 \times 18}{15}$
$AP = \frac{7 \times 18}{3} = 7 \times 6 = 42 \, cm$
Therefore,$AD = AP + PD = 42 \, cm + 18 \, cm = 60 \, cm$.

(N/A) Given: $A$ triangle $ABC$ such that $AC^2 = AB^2 + BC^2$.
To prove: $\angle B = 90^{\circ}$.
Construction: Construct a $\triangle PQR$ such that $\angle Q = 90^{\circ}$,$PQ = AB$,and $QR = BC$.
Proof:
In $\triangle PQR$,by Pythagoras theorem:
$PR^2 = PQ^2 + QR^2$
Since $PQ = AB$ and $QR = BC$,we have:
$PR^2 = AB^2 + BC^2$ ...... $(1)$
Given that $AC^2 = AB^2 + BC^2$ ...... $(2)$
From $(1)$ and $(2)$,$PR^2 = AC^2$,which implies $PR = AC$.
Now,in $\triangle ABC$ and $\triangle PQR$:
$AB = PQ$ (Construction)
$BC = QR$ (Construction)
$AC = PR$ (Proved above)
Therefore,$\triangle ABC \cong \triangle PQR$ by $SSS$ congruence criterion.
Thus,$\angle B = \angle Q$ by $CPCT$.
Since $\angle Q = 90^{\circ}$ by construction,it follows that $\angle B = 90^{\circ}$.

(N/A) Let a $\triangle ABC$ in which a line $DE$ parallel to $BC$ intersects $AB$ at $D$ and $AC$ at $E$. To prove: $DE$ divides the two sides in the same ratio,i.e.,$\frac{AD}{DB} = \frac{AE}{EC}$.
Construction: Join $BE, CD$ and draw $EF \perp AB$ and $DG \perp AC$.
Proof: Here,$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EF}{\frac{1}{2} \times DB \times EF} = \frac{AD}{DB}$ ......$(i)$ [Since,area of triangle $= \frac{1}{2} \times \text{base} \times \text{height}$]
Similarly,$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DG}{\frac{1}{2} \times EC \times DG} = \frac{AE}{EC}$ ......$(ii)$
Now,since $\triangle BDE$ and $\triangle DEC$ lie between the same parallel lines $DE$ and $BC$ and on the same base $DE$,therefore,$\text{ar}(\triangle BDE) = \text{ar}(\triangle DEC)$ ......$(iii)$
From Eqs. $(i), (ii)$ and $(iii)$,we get $\frac{AD}{DB} = \frac{AE}{EC}$.
Hence proved.