Question 9:
In the given figure, D is a point on side BC of ΔABC such that BD/CD = AB /AC. Prove that AD is the bisector of ∠BAC.
Answer:
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
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