Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that (i) ∆APB ≅ ∆AQB (ii) BP = BQ or B is equidistant from the arms ot ∠A.
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In △APB and △AQB
∠APB=∠AQB (Each 90
o
)
∠PAB=∠QAB (l is the angle bisector of ∠A)
AB=AB (Common)
∴△APB≅△AQB (By AAS congruence rule)
∴BP=BQ (By CPCT)
It can be said that B is equidistant from the arms of ∠A.
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