Question

Prove that if the equal sides of an isosceles triangle are produced, then the exterior angles so formed are obtuse and equal.

Answer 1

Consider $△ABC$ as an Isosceles triangle such that $AB=AC$ or
$\angle B=\angle C$
$180-\angle B=180-\angle C$
$ext.\angle B=ext.\angle C$
Now,sum of angles = 180
$\angle A+\angle B+\angle C=180$
$\angle B+\angle B=180-\angle A$ (Since, $\angle B=\angle C$)
$\angle B=90-\frac{1}{2}\angle A$
Thus, $\angle B$ and $\angle C$ are acute angles.
Hence, $ext.\angle B=180-\angle B$
$ext.\angle B=180-(90-\frac{1}{2}\angle A)$
$ext.\angle B=90+\frac{1}{2}\angle A$
Thus, Exterior angles are equal and obtuse.