Question

AD is an altitude of an isosceles △ ABC in which AB = AC. Show that

(i) AD bisects BC,

(ii) AD bisects ∠ A.

Answer 1

(i) Based on the △ BAD and △ CAD

We know that AD is the altitude so the angle is 90^o

So we get

∠ ADB = ∠ ADC = 90^o

It is given that AB = AC and we know that AD is common

Based on the RHS Congruence Criterion we get

△ BAD ≅ △ CAD

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.