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.