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) ∆ APB ≅ ∆ AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
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)
rms of ∠A. Or, it can be said that B is equidistant from the a
In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.