Que: 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:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Answers
Answered by
31
Solution:
It is given that the line “l” is the bisector of angle ∠A and the line segments BP and BQ are perpendiculars drawn from l.
(i) ΔAPB and ΔAQB are similar by AAS congruency because;
- ∠P = ∠Q (both are right angles)
- AB = AB (common arm)
∠BAP = ∠BAQ (As line l is the bisector of angle A)
- So, ΔAPB ≅ ΔAQB.
(ii) By the rule of CPCT, BP = BQ. So, we can say point B is equidistant from the arms of ∠A.
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Answered by
70
Answer:
In ΔAPB and ΔAQB,
∠APB = ∠AQB (Each 90º)
∠PAB = ∠QAB (l is the angle bisector of ∠A)
AB = AB (Common)
∴ ΔAPB ≅ ΔAQB (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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