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 Fig. 7.20). Show that: (i) ΔAPB ≅ ΔAQB (ii) BP = BQ or B is equidistant from the arms of ∠A.
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 (They are the two right angles) AB = AB (It is the 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, it can be said the point B is equidistant from the arms of ∠A.
