Question

IN A triangle ABC, angleB=twice of angle C and the bisector of angle B
intersects AC at D. Prove that BD/DA = BC/BA ​

Answer 1

Answer:

AΔABC ∼ ΔABD.

Thus,

AB : AD = BC : BD

In ΔBDC,

∠DBC = ∠C

⇒DC = BD ...(2) (sides opposite to equal angles are equal)

Using (2) in (1), we get

AB : AD = BC : DC

⇒ DC : AD = BC : AB

Hope it helps..