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

(i) In ΔABD and ΔACD,
∠ADB = ∠ADC [∵ Each 90°]
AB = AC [∵ Given]
AD = AD [∵ Common]
Hence, ΔABD ≅ ΔACD [∵ RHS Congruency Rule]
BD = DC [∵ CPCT]
Hence, AD bisects BC.

(ii) ∠BAD = ∠CAD [∵ CPCT]
Hence, AD bisects angle A.