Question

show that in an isosceles triangle the angles opposites to the eqal sides are eqal

Answer 1

↑ Here is your answer ↓
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Consider this triangle,

The median of an isosceles triangle is also an altitude,

In $\triangle ACD$ and $\triangle BCD$,

CD = CD {Common sides}

AD = DB {D is the midpoint}

AC = BC {Equal sides of isosceles triangle}

∴ $\triangle ACD \cong \triangle BCD$ (SSS Criterion)

=> $\angle A = \angle B$ {CPCT}

Answer 2

Let ABD be an isosceles triangle with AB = AD and AC be height.

(Refer Attachment)
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In ΔABC and ΔACD

∠ACB = ∠ACD                      (right angles)

     AB = AD                           (isosceles triangle)

     AC = AC                           (common)

By RHS rule, 
  
ΔABC congruent to ΔACD

By CPCT,

∠ABC = ∠ADC
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Hence In an isosceles triangle the angles opposites to the equal sides are

equal.
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