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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.
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