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# Line $ l $ is the bisector of an angle $ \angle \mathrm{A} $ and $ \mathrm{B} $ is any point on 1. BP and $ B Q $ are perpendiculars from $ B $ to the $ \operatorname{arms} $ of $ \angle \mathrm{A} $ (see Fig. 7.20). Show that:

**(i)** $ \triangle \mathrm{APB} \cong \triangle \mathrm{AQB} $

**(ii)** $ \mathrm{BP}=\mathrm{BQ} $ or $ \mathrm{B} $ is equidistant from the arms of $ \angle \mathrm{A} $.

"

**Given:**

Line $l$ is the bisector of an angle $\angle A$ and bisector $B$ is any point on $l$.

$BP$ and $BQ$ are perpendiculars from $B$ to the arms of $\angle A$.

**To do:**

We have to show that

(i) $\triangle APB \cong \triangle AQB$

(ii) $BP=BQ$ or $B$ is equidistant from the arms of $\angle A$.

**Solution:**

(i) We know,

Form Angle-Side- Angle:

If two angles and the included side of one triangle are equal to two angles and the included side of another triangle, then the triangles are congruent.

This implies,

$\angle P=\angle Q$ and $AB=BA$

Since the line $l$ is the bisector of $\angle A$

We get,

$\angle BAP=\angle BAQ$

Therefore,

$\triangle APB \cong \triangle AQB$.

(ii) We know that,

From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding angles and sides must be equal.

This implies,

$BP=BQ$ and $B$ is equidistant from the arms of $\angle A$.

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