Question

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

Answer 1

Let ABC be the isosceles triangle with side AB = AC.
Join vertex A to mid-point of side BC meeting BC at D..
Now in triangles ABD and ACD,
side AB = AC (Given), 
side BD = DC (Each being equal to half of BC)
side AD is common to both.
Hence triangles ABD and ACD are congruent.
Therefore angle B opposite side AC = angle C opposite side AB.
                                                   

Answer 2

prove that angles opposite to equal sides of an isosceles triangle are equal

Consider ΔABC an isosceles triangle

⇒ AB=AC

Construction: AD is bisector on BC

So, ∠BAD=∠CAD

In △ABD and △ACD

⇒ AB=AC

⇒ ∠BAD=∠CAD

⇒ AD=AD

So, △ABD≅△ACD ..........(SAS)

hence, ∠ABC=∠ACB