# 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 $\angle \mathrm{A}$."

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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 $\angle A$.

Updated on 10-Oct-2022 13:40:53