Question

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

(i) AD bisects BC

(ii) AD bisects ∠A.

Answer 1

(i) In ∆BAD and ∆CAD,

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

AB = AC (Given)

AD = AD (Common)

∠∆BAD ≅ ∠∆CAD (By RHS Congruence rule)

BD = CD (By CPCT)

Hence, AD bisects BC.

(ii) Also, by CPCT,

∠BAD = ∠CAD

Hence, ∠AD bisects A.