2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects ∠A.
(i) In ΔABD and ΔACD, ∠ADB = ∠ADC = 90° AB = AC (It is given in the question) AD = AD (Common arm) ∴ ΔABD ≅ ΔACD by RHS congruence condition. Now, by the rule of CPCT, BD = CD. So, AD bisects BC (ii) Again, by the rule of CPCT, ∠BAD = ∠CAD Hence, AD bisects ∠A
