Question

In ΔABC, AB = AC. D is a point in the interior of the triangle such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of ΔABC.

Answer 1

Since AB = AC

∠ABC = ∠ACB

But ∠DBC = ∠DBC

⇒ ∠ABD = ∠ACD

Now in ΔABD and ΔADC

AB = AC

AD = AD

∠ABD = ∠ACD

Therefore, ΔABD ≅ ΔADC  ...(SSA criteria)

Hence, ∠BAD = ∠CAD

Thus, AD bisects ∠BAC.