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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.
Solution
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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