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 $\angle A$.
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