# 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:**

Given: l is the bisector of an angle ∠A and BP ⊥ AP and BQ ⊥ AQ

To Prove: ΔAPB ≅ ΔAQB and BP = BQ

i) We can show two triangles APB and AQB are congruent by using AAS congruency rule and then show that the corresponding parts of congruent triangles will be equal.

In ΔAPB and ΔAQB,

∠BAP = ∠BAQ (l is the angle bisector of ∠A)

∠APB = ∠AQB (Each 90°)

AB = AB (Common)

∴ ΔAPB ≅ ΔAQB (By AAS congruence rule)

ii) Since, ΔAPB ≅ ΔAQB

∴ BP = BQ (By CPCT)

Or, it can be said that point B is equidistant from the arms of ∠A.

**Video Solution:**

## 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

### NCERT Maths Solutions Class 9 - Chapter 7 Exercise 7.1 Question 5:

**Summary:**

If line L is the bisector of an angle ∠A and ∠B is any point on l, BP and BQ are perpendiculars from B the arms of ∠A, then ΔAPB ≅ ΔAQB using AAS congruence and BP = BQ or B is equidistant from the arms of ∠A using CPCT.