Question 5:
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:
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Answer:
In ∆APB and ∆AQB,
∠APB = ∠AQB (Each 90º)
∠PAB = ∠QAB (l is the angle bisector of ∠A)
AB = AB (Common)
∆APB ≅ ∆AQB (By AAS congruence rule)
∴ BP = BQ (By CPCT)
Or, it can be said that B is equidistant from the arms of ∠A.
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