# Chapter-5: Congruence of Triangles

Question 1:

In ΔABC$\Delta ABC$, if AB = AC and $\angle$A = 70o, find $\angle$B and $\angle$C.

Solution:

AB = AC implies their opposite angles are equal

B=C$\Rightarrow \angle B=\angle C$ [angles opposites to equal sides]

But A+B+C=180$\angle A+\angle B+\angle C=180^{\circ}$

70+B+B=180$\Rightarrow 70^{\circ}+\angle B+\angle B=180^{\circ}$ 70+2B=180$\Rightarrow 70^{\circ}+2\angle B=180^{\circ}$ 2B=18070$\Rightarrow 2\angle B=180^{\circ}-70^{\circ}$ 2B=110$\Rightarrow 2\angle B=110^{\circ}$ B=1102$\Rightarrow \angle B=\frac{110^{\circ}}{2}$ B=55$\Rightarrow \angle B=55^{\circ}$ B=C=55$\Rightarrow \angle B=\angle C=55^{\circ}$

Question 2:

The vertical angle of an isosceles triangle is 100o. Find its base angles.

Solution:

Consider the isosceles triangle ΔABC$\Delta ABC$.

Since the vertical angle of ABC is 100o, we have, A$\angle A$ = 100o

By angle sum property of a triangle, we have,

A+B+C=180$\angle A+\angle B+\angle C=180^{\circ}$ 100+B+C=180$\Rightarrow 100^{\circ}+\angle B+\angle C=180^{\circ}$

100+2B=180$\Rightarrow 100^{\circ}+2\angle B=180^{\circ}$ [since in an isosceles triangle base angles are equal, B=C$\angle B=\angle C$]

2B=180100=80$\Rightarrow 2\angle B=180^{\circ}-100^{\circ}=80^{\circ}$ B=802$\Rightarrow \angle B=\frac{80^{\circ}}{2}$ B=40$\Rightarrow \angle B=40^{\circ}$ B=C=40$\Rightarrow \angle B=\angle C=40^{\circ}$

Question 3:

In a Δ$\Delta$ABC, if AB = AC and $\angle$B = 65o, find $\angle$C and $\angle$A.

Solution:

In ΔABC$\Delta ABC$, if AB = AC

ΔABC$\Rightarrow \Delta ABC$ is an isosceles triangle

$\Rightarrow$ Base angles are equals

B=C$\Rightarrow \angle B=\angle C$

C=65$\Rightarrow \angle C=65^{\circ}$ [Since B=65$\angle B=65^{\circ}$]

Also by angle sum property, we have

A+B+C=180$\angle A+\angle B+\angle C=180^{\circ}$

A+65+65=180$\Rightarrow \angle A+65^{\circ}+65^{\circ}=180^{\circ}$ $\angle B=\angle C=65^{\circ}$

A=180130=50$\Rightarrow \angle A=180^{\circ}-130^{\circ}=50^{\circ}$

Question 4:

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

Solution:

Let ABC be an isosceles triangle in which AB = AC.

Then we have B=C$\angle B=\angle C$

Let B=C=x$\angle B=\angle C=x$

Then vertex angle A = 2(x + x) = 4x

Now, x + x + 4x = 180o

6x=180$\Rightarrow 6x=180^{\circ}$ x=1806=30$\Rightarrow x=\frac{180^{\circ}}{6}=30^{\circ}$

Therefore, Vertex A=4×30=120$\angle A=4\times 30=120^{\circ}$

And, B=C=30$\angle B=\angle C=30^{\circ}$

Question 5:

What is the measure of each of the equal angles of a right-angled isosceles triangle?

Solution:

In a right angle isosceles triangle, the vertex angle is A=90$\angle A=90^{\circ}$ and the other two base angles are equal.

Let xo be the base angle and we have, B=C=x$\angle B=\angle C=x$.

By angle sum property of a triangle, we have

A+B+C=180$\angle A+\angle B+\angle C=180^{\circ}$ 90+x+x=180$\Rightarrow 90^{\circ}+x+x=180^{\circ}$ 90+2x=180$\Rightarrow 90^{\circ}+2x=180^{\circ}$ 2x=18090=90$\Rightarrow 2x=180^{\circ}-90^{\circ}=90^{\circ}$ x=902=45$\Rightarrow x=\frac{90^{\circ}}{2}=45^{\circ}$

Thus, we have B=C=45$\angle B=\angle C=45^{\circ}$

Question 6:

If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.

Solution:

Given: ABC is an isosceles triangle in which AB = AC and BC is produced both ways

To Prove: $\angle$EBA = $\angle$DCA

Proof: In Δ$\Delta$ABC we have,

AB = AC

$\Rightarrow$ $\angle$B = $\angle$C

Now exterior $\angle$EBA = $\angle$A + $\angle$C = $\angle$A + $\angle$B [$\angle$B = $\angle$C]

and exterior $\angle$DCA = $\angle$A + $\angle$B

$\Rightarrow$ Exterior $\angle$EBA = Exterior $\angle$DCA.

Question 7:

Find the measure of each exterior angle of an equilateral triangle.

Solution:

Let ΔABC$\Delta ABC$ be an equilateral triangle.

Since it is an equilateral triangle, all the angle is 60o

The exterior angle of A$\angle A$ = BAF$\angle BAF$

The exterior angle of B$\angle B$ = ABD$\angle ABD$

The exterior angle of C$\angle C$ = ACE$\angle ACE$

We can observe that angles A$\angle A$ and BAF$\angle BAF$, B$\angle B$ and ABD$\angle ABD$, C$\angle C$ and ACE$\angle ACE$ form linear pairs.

Therefore, we have

A+BAF=180$\angle A+\angle BAF=180^{\circ}$ 60+BAF=180$\Rightarrow 60^{\circ}+\angle BAF=180^{\circ}$ BAF=18060=120$\Rightarrow \angle BAF=180^{\circ}-60^{\circ}=120^{\circ}$

Similarly, we have

B+ABD=180$\angle B+\angle ABD=180^{\circ}$ 60+ABD=180$\Rightarrow 60^{\circ}+\angle ABD=180^{\circ}$ ABD=18060=120$\Rightarrow \angle ABD=180^{\circ}-60^{\circ}=120^{\circ}$

Also, we have

C+ACE=180$\angle C+\angle ACE=180^{\circ}$ 60+ACE=180$\Rightarrow 60^{\circ}+\angle ACE=180^{\circ}$ ACE=18060=120$\Rightarrow \angle ACE=180^{\circ}-60^{\circ}=120^{\circ}$

Thus, we have, BAF=120$\angle BAF=120^{\circ}$, ABD=120$\angle ABD=120^{\circ}$, ACE=120$\angle ACE=120^{\circ}$

So, the measure of each exterior angle of an equilateral triangle is 120o.

Question 8:

In the given figure, O is the mid-point of each of the line segments AB and CD. Prove that AC = BD and AC || BD.

Solution:

Given: Two lines AB and CD intersect at O and O is the midpoint of AB and CD.

$\Rightarrow$ AO = OB and CO = OD

To prove: AC = BD and AC || BD

Proof:

In Δ$\Delta$AOC and Δ$\Delta$BOD, we have,

AO = OB [Given: O is the midpoint of AB]

$\angle$AOC = $\angle$BOD [Vertically opposite angles]

CO = OD [Given: O is the midpoint of CD]

So, by Side-Angle-Side congruence triangle are equal.

Therefore, we have, AC = BD.

Similarly, by c.p.c.t, we have, $\angle$ACO = $\angle$BDO and $\angle$CAO = $\angle$DBO

This implies that alternate angles formed by AC and BD with transversal CD are equal. This means that, AC || BD. Thus, AC = BD and AC || BD.