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Question

$$ABC$$ is an isosceles triangle with $$AB = AC$$. Show that $$\angle B = \angle C$$. 
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Solution

Given,
$$ABC$$ is an isosceles triangle with $$AB = AC$$. 
To Prove : $$\angle B = \angle C$$ 
Construction: Draw $$AP \perp BC$$.
 
Proof: In $$\Delta ABC, AP \perp BC$$ and $$AB = BC$$. 
$$\therefore$$ In $$\Delta ABP$$ and $$\Delta ACP$$
$$\angle APB = \angle APC = 90^0 (\because AP \perp BC)$$ 
Hypotenuse $$AB$$ = Hypotenuse $$AC$$ 
$$AP$$ is common. 
As per $$RHS$$ postulate, 
$$\Delta ABP \cong \Delta ACP$$ 
$$\therefore \ \angle ABP = \angle ACP$$  [by CPCT]
$$\therefore \ \ \angle ABC = \angle ACB$$ 
$$\therefore \ \angle B = \angle C$$.   $$(Hence\  proved)$$

Mathematics

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