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# 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.

**☛ Check: **NCERT Solutions for Class 9 Maths Chapter 7

**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.

**☛ Related Questions:**

- ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show thati) ΔABD ≅ ΔACDii) ΔABP ≅ ΔACPiii) AP bisects ∠A as well as ∠Div) AP is the perpendicular bisector of BC.
- Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see Fig. 7.40). Show that:(i) Δ ABM ≅ Δ PQN(ii) Δ ABC ≅ Δ PQR
- BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
- ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

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