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In the given figure, 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. Show that B is equidistant from the arms of ∠A.

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In Δs APB and ABQ, we have

 ∠APB = ∠AQB           (Each = 90°)

∠PAB = ∠QAB           (AB bisects ∠PAQ ) 

AB = BA             (common) 

∴ ΔAPB ≅ ABQ      (AAS)

⇒ BP = BQ       (cpct)

⇒ B is equidistant from the arms of ∠A.

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