Question

5. Linel is the bisector of an angle Z A and B is any
point on I. BP and BQ are perpendiculars from B
to the arms of Z A (see Fig. 7.20). Show that:
(i) A APB=AAQB
(ii) BP = BQ or B is equidistant from the arms
of ZA​

Answer 1

Answer: 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) ΔAPB ≅ ΔAQB (ii) BP = BQ or B is equidistant from the arms of ∠A. Given: l ... BP & BQ are perpendiculars from B, So, APB = AQB = 90 To prove: (i) APB AQB (ii)  

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