In Figure, D is a point on side BC of ∆ ABC such that $$\dfrac{BD}{CD}=\dfrac{AB}{AC}$$. Prove that AD is the bisector of ∠ BAC.

Asked by Pragya Singh | 1 year ago |  207

##### Solution :-

In the given let us extend BA to P such that;

AP = AC.

Now join PC.

Given, $$\dfrac{BD}{CD}=\dfrac{AB}{AC}$$

$$\dfrac{BD}{CD}=\dfrac{AP}{AC}$$

By using the converse of basic proportionality theorem, we get,

∠BAD = ∠APC (Corresponding angles) ……….. (i)

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

By the new figure, we have;

AP = AC

∠APC = ∠ACP ………………. (iii)

On comparing equations (i), (ii), and (iii), we get,

Therefore, AD is the bisector of the angle BAC.

Hence, proved.

Answered by Abhisek | 1 year ago

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