French, asked by athlete01, 11 months ago

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 Anonymous
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.

______________________

Hope it will be helpful :)

Answered by Anonymous
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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