# AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC (ii) AD bisects ∠A

**Solution:**

Given: AB = AC

Let's construct an isosceles triangle ABC in which AB = AC as shown below.

**(i)** In ΔBAD and ΔCAD,

∠ADB = ∠ADC (Each 90° as AD is an altitude)

AB = AC (Given)

AD = AD (Common)

∴ ΔBAD ≅ ΔCAD (By RHS Congruence rule)

∴ BD = CD (By CPCT)

Hence, AD bisects BC.

**(ii)** Since, ΔBAD ≅ ΔCAD

By CPCT, ∠BAD = ∠CAD

Hence, AD bisects ∠A.

**Video Solution:**

## AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects ∠A

### NCERT Maths Solutions Class 9 - Chapter 7 Exercise 7.3 Question 2:

**Summary:**

If AD is an altitude of an isosceles triangle ABC in which AB = AC, then AD bisects BC and AD bisects ∠A.