Question

using the exterior angle inequality prove the base angles of any isoceles triangle are acute

Answer 1

Let angle ABC=x

Then angle ACB=x (base angles of isosceles∆ are equal)

This implies that ACD= 180°-x (angles in a straight line add up to 180°)

This implies that angle ABC<ACD and angle ACB<ACD.

This implies that x<180°-x

= x+x<180°

=2x<180°

=x<90°

Therefore, angle ABC<90° and angle ACB<90°