AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC (ii) AD bisects ∠A.
Given,
AD is an altitude and AB = AC
(i) In ΔABD and ΔACD,
∠ADB = ∠ADC = 90°
AB = AC (Given)
AD = AD (Common)
Therefore, ΔABD ≅ ΔACD by RHS congruence condition.
Now,
BD = CD (by CPCT)
Thus, AD bisects BC
(ii) ∠BAD = ∠CAD (by CPCT)
Thus, AD bisects ∠A.
