Question

In the given figure, D is a point on side BC of ΔABC such that ∠ADC=∠BAC . Prove that AD is the bisector of ∠BAC.

Answer 1

Let us extend BA to P such that AP = AC. Join PC.

It is given that,

(BD)/(CD) = (AB)/(AC)

=> "BD"/"CD" = "AP"/"AC"

By using the converse of basic proportionality theorem, we obtain

AD || PC

⇒ ∠BAD = ∠APC (Corresponding angles) … (1)

And, ∠DAC = ∠ACP (Alternate interior angles) … (2)

By construction, we have

AP = AC

⇒ ∠APC = ∠ACP … (3)

On comparing equations (1), (2), and (3), we obtain

∠BAD = ∠APC

⇒ AD is the bisector of the angle BAC.