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

(i) Based on the ΔBAD and ΔCAD
We know that AD is the altitude so the angle is 90°.
So we get ∠ADB = ∠ADC = 90°
It is given that AB=AC and we know that AD is common
Based on the RHS Congruence Criterion we get ΔBAD ≅ ΔCAD
So we get BD=CD (c.p.c.t)
Therefore, it is proved that AD bisects BC
(ii) We also know that ∠BAD = ∠CAD (c.p.c.t)
Therefore, it is proved that AD bisects ∠A.