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Question
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:
- ΔAPB ≅ ΔAQB
- BP = BQ or B is equidistant from the arms of ∠A.
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Solution
We have, l is the bisector of ∠QAP.
∴ ∠QAB = ∠PAB
and ∠Q = ∠P ...[Each 90°]
⇒ ∠ABQ = ∠ABP ...[By angle sum property of △]
i. Now, in △APB and △AQB, we have
∠ABP = ∠ABQ ...[Proved above]
AB = BA ...[Common]
∠PAB ≅ ∠QAB ...[Given]
△APB ≅ △AQB ...[By ASA congruency]
ii. Since, △APB ≅ △AQB
⇒ BP = BQ ...[By Corresponding parts of congruent triangles]
i.e., [Perpendicular distance of B from AP] = [Perpendicular distance of B from AQ]
Thus, the point B is equidistant from the arms of ∠A.
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