Question

Prove that the angles opposite to equal sides of the isosceles triangle are equal.

Answer 1

Hint: First of all draw an angle bisector from angle A to the opposite side from angle A. After drawing an angle bisector, the triangles are divided into two triangles then prove that the two triangles are congruent and after proving the congruence of two triangles from CPCT we can show that the angles opposite to equal sides are equal.

Complete step-by-step answer:
In the below figure, we have drawn an isosceles triangle ABC and AD is an angle bisector from A to BC.

From the above figure, as AD is the angles bisector from A to BC then$\angle BAC=\angle CAD$.
In$\Delta BAD\text{ and }\Delta \text{CAD}$,
It is given that AB is equal to AC, the property of isosceles triangle.
AB = AC
AD is the angle bisector of A so,
$\angle BAD=\angle CAD$
AD is common in both the triangles.
So, from SAS congruence we can say that triangle BAD is congruent to triangle CAD and mathematically, it is written as $\Delta BAD\cong \Delta CAD$.
Now,$\Delta BAD\cong \Delta CAD$ then from the CPCT (Corresponding part of the congruent triangles):$\angle ABD=\angle ACD$
Or, we can say that$\angle B=\angle C$.
As AB = AC and the relation between angles opposite to them are$\angle B=\angle C$
From the above statement we have established that the angles opposite to equal sides of an isosceles $\Delta ABC$are equal.
Hence, we have proved that angles opposite to equal sides of an isosceles triangle are equal.

Note: As we know that all the sides of the equilateral triangle are equal and angles opposite to these sides are also equal so from this property of the equilateral triangle we can say that angles opposite to equal sides of the isosceles triangle are also equal.