Congruence of Triangles Questions in English

(N/A) $(i)$ In $\Delta AOD$ and $\Delta BOC$:
$OA = OB$ (Given)
$OD = OC$ (Given)
Also,$\angle AOD = \angle BOC$ (Vertically opposite angles).
Therefore,by the $SAS$ congruence rule,$\Delta AOD \cong \Delta BOC$.
$(ii)$ Since $\Delta AOD \cong \Delta BOC$,their corresponding parts are equal $(CPCT)$.
Thus,$\angle OAD = \angle OBC$.
These are alternate interior angles formed by the transversal $AB$ intersecting lines $AD$ and $BC$.
Since the alternate interior angles are equal,$AD \parallel BC$.

(N/A) Line $l$ is the perpendicular bisector of $AB$ and passes through $C$,which is the mid-point of $AB$ (see figure).
We need to show that $PA = PB$.
Consider $\Delta PCA$ and $\Delta PCB$.
In these triangles:
$AC = BC$ ($C$ is the mid-point of $AB$)
$\angle PCA = \angle PCB = 90^\circ$ (Given,as $l$ is the perpendicular bisector)
$PC = PC$ (Common side)
Therefore,by the $SAS$ congruence rule,$\Delta PCA \cong \Delta PCB$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$PA = PB$,which means $P$ is equidistant from $A$ and $B$.

(N/A) In quadrilateral $ACBD$,we are given:
$AC = AD$
$AB$ bisects $\angle A$,which implies $\angle CAB = \angle DAB$.
Now,consider $\Delta ABC$ and $\Delta ABD$:
$1$. $AC = AD$ (Given)
$2$. $\angle CAB = \angle DAB$ (Since $AB$ bisects $\angle A$)
$3$. $AB = AB$ (Common side)
By the $SAS$ (Side-Angle-Side) congruence criterion,$\Delta ABC \cong \Delta ABD$.
Since the triangles are congruent,their corresponding parts are equal by $CPCT$ (Corresponding Parts of Congruent Triangles).
Therefore,$BC = BD$.

(N/A) Given: $AD \perp AB$,$BC \perp AB$ and $AD = BC$.
To prove: $CD$ bisects $AB$,i.e.,$OA = OB$.
Proof:
In $\Delta OBC$ and $\Delta OAD$:
$1$. $\angle OBC = \angle OAD = 90^\circ$ (Given)
$2$. $\angle BOC = \angle AOD$ (Vertically opposite angles)
$3$. $BC = AD$ (Given)
By $AAS$ congruence rule,$\Delta OBC \cong \Delta OAD$.
Therefore,$OB = OA$ (by $c.p.c.t.$).
Since $OB = OA$,$O$ is the midpoint of $AB$.
Hence,$CD$ bisects $AB$.

(N/A) Given: Line $l$ is the bisector of $\angle A$. $BP \perp AQ$ and $BQ \perp AP$.
$(i)$ In $\Delta APB$ and $\Delta AQB$:
$\angle APB = \angle AQB = 90^{\circ}$ (Given)
$\angle PAB = \angle QAB$ ($l$ is the bisector of $\angle A$)
$AB = AB$ (Common side)
Therefore,by $AAS$ congruence criterion,$\Delta APB \cong \Delta AQB$.
$(ii)$ Since $\Delta APB \cong \Delta AQB$,their corresponding parts are equal by $CPCT$.
Therefore,$BP = BQ$.
This shows that point $B$ is equidistant from the arms of $\angle A$.

(N/A) We have $\angle BAD = \angle EAC$.
Adding $\angle DAC$ on both sides,we have:
$\angle BAD + \angle DAC = \angle EAC + \angle DAC$
$\Rightarrow \angle BAC = \angle DAE$
Now,in $\triangle ABC$ and $\triangle ADE$,we have:
$\angle BAC = \angle DAE$ [Proved above]
$AB = AD$ [Given]
$AC = AE$ [Given]
$\therefore \triangle ABC \cong \triangle ADE$ [Using $SAS$ congruence criterion]
Since $\triangle ABC \cong \triangle ADE$,therefore,their corresponding parts are equal $(CPCT)$.
$\Rightarrow BC = DE$.

(N/A) Given: $P$ is the mid-point of $AB$.
$\therefore AP = BP$
$\angle EPA = \angle DPB$ [Given]
Adding $\angle EPD$ to both sides,we get:
$\angle EPA + \angle EPD = \angle DPB + \angle EPD$
$\Rightarrow \angle APD = \angle BPE$
$(i)$ In $\Delta DAP$ and $\Delta EBP$:
$AP = BP$ [Proved above]
$\angle PAD = \angle PBE$ [Given as $\angle BAD = \angle ABE$]
$\angle APD = \angle BPE$ [Proved above]
By $ASA$ congruence criterion,we have:
$\Delta DAP \cong \Delta EBP$
$(ii)$ Since $\Delta DAP \cong \Delta EBP$,their corresponding parts are equal $(CPCT)$:
$\Rightarrow AD = BE$

(N/A) Given: In $\Delta ABC$,$\angle C = 90^\circ$ and $M$ is the mid-point of $AB$ $(AM = BM)$. $CM$ is produced to $D$ such that $DM = CM$.
$(i)$ In $\Delta AMC$ and $\Delta BMD$:
$AM = BM$ (Given,$M$ is the mid-point of $AB$)
$\angle AMC = \angle BMD$ (Vertically opposite angles)
$CM = DM$ (Given)
Therefore,by $SAS$ congruence criterion,$\Delta AMC \cong \Delta BMD$.
$(ii)$ Since $\Delta AMC \cong \Delta BMD$,their corresponding parts are equal $(c.p.c.t.)$.
So,$\angle MAC = \angle MBD$. These are alternate interior angles,which implies $AC \parallel DB$.
Since $AC \parallel DB$ and $BC$ is a transversal,the sum of consecutive interior angles is $180^\circ$.
$\angle ACB + \angle DBC = 180^\circ$
$90^\circ + \angle DBC = 180^\circ$
$\angle DBC = 180^\circ - 90^\circ = 90^\circ$.
$(iii)$ In $\Delta DBC$ and $\Delta ACB$:
$DB = AC$ ($c.p.c.t.$ from $\Delta AMC \cong \Delta BMD$)
$\angle DBC = \angle ACB = 90^\circ$ (Proved above)
$BC = CB$ (Common side)
Therefore,by $SAS$ congruence criterion,$\Delta DBC \cong \Delta ACB$.
$(iv)$ Since $\Delta DBC \cong \Delta ACB$,their corresponding parts are equal $(c.p.c.t.)$.
So,$DC = AB$.
Since $DM = CM$,$CM = \frac{1}{2} DC$.
Substituting $DC = AB$,we get $CM = \frac{1}{2} AB$.

(N/A) $(i)$ In $\Delta ABC$,we have
$AB = AC$ [Given]
$\therefore \angle C = \angle B$ [Angles opposite to equal sides are equal]
$\Rightarrow \frac{1}{2} \angle C = \frac{1}{2} \angle B$
Or $\angle OCB = \angle OBC$
$\Rightarrow OB = OC$ [Sides opposite to equal angles are equal]
$(ii)$ In $\Delta ABO$ and $\Delta ACO$,we have
$AB = AC$ [Given]
$OB = OC$ [Proved above]
$\angle OBA = \angle OCA$ [Since $\frac{1}{2} \angle B = \frac{1}{2} \angle C$]
$\therefore$ Using $SAS$ congruence criteria,
$\Delta ABO \cong \Delta ACO$
$\Rightarrow \angle OAB = \angle OAC$ [c.p.c.t.]
$\Rightarrow AO$ bisects $\angle A$.

(N/A) $(i)$ In $\Delta ABE$ and $\Delta ACF$,we have:
$\angle AEB = \angle AFC = 90^\circ$ (Since $BE \perp AC$ and $CF \perp AB$)
$\angle A = \angle A$ (Common angle)
$BE = CF$ (Given)
Therefore,by the $AAS$ congruence criterion,$\Delta ABE \cong \Delta ACF$.
$(ii)$ Since $\Delta ABE \cong \Delta ACF$,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$.
Thus,$ABC$ is an isosceles triangle.

(N/A) $(i)$ In $\Delta ABD$ and $\Delta ACD$,we have:
$AB = AC$ [Given]
$AD = AD$ [Common]
$BD = CD$ [Given]
$\therefore \Delta ABD \cong \Delta ACD$ [$SSS$ congruence criterion]
$(ii)$ In $\Delta ABP$ and $\Delta ACP$,we have:
$AB = AC$ [Given]
$\angle BAP = \angle CAP$ [Since $\Delta ABD \cong \Delta ACD$,their corresponding parts are equal]
$AP = AP$ [Common]
$\therefore \Delta ABP \cong \Delta ACP$ [$SAS$ congruence criterion]
$(iii)$ Since $\Delta ABP \cong \Delta ACP$,their corresponding parts are congruent.
$\Rightarrow \angle BAP = \angle CAP$,so $AP$ bisects $\angle A$.
Also,in $\Delta BDP$ and $\Delta CDP$:
$BD = CD$ [Given]
$DP = DP$ [Common]
$BP = CP$ [Since $\Delta ABP \cong \Delta ACP$,$BP = CP$ by $CPCT$]
$\therefore \Delta BDP \cong \Delta CDP$ [$SSS$ congruence criterion]
$\Rightarrow \angle BDP = \angle CDP$,so $AP$ bisects $\angle D$.
$(iv)$ Since $\Delta ABP \cong \Delta ACP$,we have $\angle APB = \angle APC$ [by $CPCT$].
Since $\angle APB + \angle APC = 180^\circ$ [Linear pair],
$\angle APB = \angle APC = 90^\circ$.
Also $BP = CP$ [by $CPCT$].
Thus,$AP$ is the perpendicular bisector of $BC$.

(B) To find the number of equilateral triangles of side $1\,cm$ that can fit into the given shapes,we calculate the area of the shapes and divide by the area of one small equilateral triangle.
$1$. For the regular hexagon of side $s = 5\,cm$:
Area $= \frac{3\sqrt{3}}{2} s^2 = \frac{3\sqrt{3}}{2} \times 25 = 37.5\sqrt{3} \approx 64.95\,cm^2$.
Each small triangle of side $1\,cm$ has an area of $\frac{\sqrt{3}}{4} \times 1^2 = 0.25\sqrt{3} \approx 0.433\,cm^2$.
Number of triangles in Fig. $(i) = \frac{37.5\sqrt{3}}{0.25\sqrt{3}} = 150$.
$2$. For the star-shaped figure in Fig. $(ii)$:
This figure consists of a central hexagon of side $5\,cm$ and $6$ equilateral triangles of side $5\,cm$ attached to its sides.
Total area $= \text{Area of hexagon} + 6 \times \text{Area of triangle of side } 5\,cm$.
Total area $= 37.5\sqrt{3} + 6 \times (6.25\sqrt{3}) = 37.5\sqrt{3} + 37.5\sqrt{3} = 75\sqrt{3} \approx 129.9\,cm^2$.
Number of triangles in Fig. $(ii) = \frac{75\sqrt{3}}{0.25\sqrt{3}} = 300$.
Therefore,the star-shaped figure $(ii)$ has more triangles.

(D) Given that $\triangle ABC \cong \triangle FDE$.
$1$. In $\triangle ABC$,the sum of angles is $180^{\circ}$. Therefore,$\angle C = 180^{\circ} - (80^{\circ} + 40^{\circ}) = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
$2$. Since $\triangle ABC \cong \triangle FDE$,the corresponding parts of congruent triangles are equal $(CPCT)$.
$3$. The correspondence is $A \leftrightarrow F$,$B \leftrightarrow D$,and $C \leftrightarrow E$.
$4$. Therefore,$AB$ corresponds to $FD$ (or $DF$). Thus,$DF = AB = 5 \, cm$.
$5$. Also,$\angle C$ corresponds to $\angle E$. Thus,$\angle E = \angle C = 60^{\circ}$.
$6$. Comparing this with the given options,$DF = 5 \, cm$ and $\angle E = 60^{\circ}$ is correct.

(B) It is false to say that $BC = QR$.
Since $\triangle ABC \cong \triangle RPQ$,the corresponding parts of congruent triangles are equal $(CPCT)$.
By matching the vertices,we have $A \leftrightarrow R$,$B \leftrightarrow P$,and $C \leftrightarrow Q$.
Therefore,the side $BC$ corresponds to the side $PQ$.
Thus,$BC = PQ$,not $QR$.

(A) Yes,it is true that $PR = EF$.
Since $\Delta PQR \cong \Delta EDF$,the corresponding parts of congruent triangles are equal $(CPCT)$.
The vertices correspond as follows: $P \leftrightarrow E$,$Q \leftrightarrow D$,and $R \leftrightarrow F$.
Therefore,the side $PR$ corresponds to the side $EF$,which implies $PR = EF$.

(A) Let the vertices of the first triangle be $A, B, C$ and the vertices of the second triangle be $P, Q, R$.
$A$ one-one correspondence is a mapping between the sets of vertices of the two triangles.
The number of ways to map $3$ vertices of the first triangle to $3$ vertices of the second triangle is given by the number of permutations of $3$ elements,which is $3!$.
$3! = 3 \times 2 \times 1 = 6$.
Therefore,there are $6$ possible one-one correspondences between two triangles.

(B) In the given correspondence $ABC \leftrightarrow PRQ$,the vertices correspond as follows:
$A$ corresponds to $P$
$B$ corresponds to $R$
$C$ corresponds to $Q$
Therefore,the side $AB$ corresponds to the side formed by the corresponding vertices of $A$ and $B$,which is $PR$.

(C) Given that $\Delta ABC \cong \Delta YZX$.
By the property of congruent triangles,the corresponding parts of congruent triangles are equal $(CPCT)$.
Comparing the vertices: $A$ corresponds to $Y$,$B$ corresponds to $Z$,and $C$ corresponds to $X$.
Therefore,side $AB$ corresponds to side $YZ$.
Thus,side $AB = YZ$.

(A) Given that $\Delta ABC \cong \Delta RPQ$.
By the property of congruent triangles,corresponding parts of congruent triangles are equal $(CPCT)$.
This implies that the sides of $\Delta ABC$ are equal to the corresponding sides of $\Delta RPQ$.
Specifically,$AB = RP$,$BC = PQ$,and $AC = RQ$.
The perimeter of a triangle is the sum of its three sides.
Perimeter of $\Delta ABC = AB + BC + AC = 18 \, cm$.
Since $\Delta ABC \cong \Delta RPQ$,the perimeter of $\Delta RPQ$ is equal to the perimeter of $\Delta ABC$.
Perimeter of $\Delta RPQ = RP + PQ + RQ = AB + BC + AC = 18 \, cm$.
Note that the perimeter of $\Delta PQR$ is the same as the perimeter of $\Delta RPQ$ because the set of side lengths is identical.
Therefore,the perimeter of $\Delta PQR$ is $18 \, cm$.

(C) Two triangles are said to be congruent if they are identical in shape and size.
If $\Delta ABC \cong \Delta DEF$,it implies that all corresponding sides and angles are equal.
As a consequence of congruence,the areas of two congruent triangles must be equal.
Given that the area of $\Delta ABC = 58 \ cm^2$.
Therefore,the area of $\Delta DEF = 58 \ cm^2$.

(C) Given that $\Delta ABC \cong \Delta RPQ$.
By the property of congruent triangles,the corresponding parts of congruent triangles are equal $(CPCT)$.
The vertices correspond as follows: $A \leftrightarrow R$,$B \leftrightarrow P$,and $C \leftrightarrow Q$.
Therefore,the side $AB$ of $\Delta ABC$ corresponds to the side $RP$ of $\Delta RPQ$.
Since $\Delta RPQ$ is the same as $\Delta PQR$ (as the set of vertices is the same),the side $RP$ is the same as side $PR$ in $\Delta PQR$.
Thus,side $AB$ is equal to side $RP$.