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.