Question

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

(i) AD bisects BC (ii) AD bisects ∠A.

Answer 1

(i) In ΔBAD and ΔCAD,

∠ADB = ∠ADC (Each 90º 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) Also, by CPCT,

∠BAD = ∠CAD

Hence, AD bisects ∠A.