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
Section | Number of girls per thousand boys |
|---|---|
Scheduled Caste (SC) | 940 |
Scheduled Tribe (ST) | 970 |
Non-SC/ST | 920 |
Backward districts | 950 |
Non-backward districts | 920 |
Rural | 930 |
Urban | 910 |
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
