Answer :
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.
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.
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