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Question

In the figure, $$ABC$$ is a triangle in which $$AD$$ is the bisector of $$\angle A$$. If $$AD\bot BC$$, show that $$\triangle ABC$$ is isosceles.

Solution
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Given:
$$AD$$ is the bisector of $$\angle A$$
$$\Rightarrow\angle DAB=\angle DAC$$ ...$$(1)$$
$$AD\bot BC$$
$$\Rightarrow\angle BDA=\angle CDA={90}^{o}$$
To prove:
$$\triangle ABC$$ is isosceles.
Proof:
In $$\triangle DAB$$ and $$\triangle DAC$$,
$$\angle BDA=\angle CDA={90}^{o}$$
$$DA=DA$$ [common]
$$\angle DAB=\angle DAC$$ [from $$(1)$$]
$$\therefore\triangle DAB\cong \triangle DAC$$ [By ASA congruence property]
$$\Rightarrow AB=AC$$
Hence, $$\triangle ABC$$ is isosceles

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