$Δ A B C$ is an isosceles triangle with $A B = A C$ . Draw $A P ⊥ B C$ to show that $∠ B = ∠ C$ .

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Solution

$A B C$ is an isosceles triangle in which $A B = A C$
Draw $A P ⊥ B C$
In $Δ A B P$ and $Δ A C P$
hyp. $A B$ = hyp. $A C$ (given)
$A P = A P$ (common)
and $∠ A P B = ∠ A P C = 90^{0} (∵ A P ⊥ B C)$
$∴ Δ A B P = Δ A C P$
(by RHS congruency rule)
$∠ B = ∠ C$ (by c.p.c.t)
Hence proved.

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