Question

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

Answer 1

Given : In an isosceles $\Delta ABC,AB=AC$ and base BC is produced both ways

To prove : $\angle ACD=\angle ABE$

Proof : In $\Delta ABC,$

$\because AB=AC$

$\therefore \angle C=\angle B$ (Angles opposite to equal sides)

$\Rightarrow \angle ACB=\angle ABC$

But $\angle ACD=\angle ACB={180}^{\circ }$ (Linear pair)

and $\angle ABE+\angle ABC={180}^{\circ }$

$\therefore \angle ACD+\angle ACB=\angle ABE+\angle ABC$

But $\angle ACB=\angle ABC$ (Proved)

$\therefore \angle ACD=\angle ABE$

Hence proved.