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

In the given figure, let us extend BA to P such that;
AP = AC.
Now join PC.

Given, $$\frac{BD}{CD} = \frac{AB}{AC}$$
$$\frac{ BD}{CD} = \frac{AP}{AC}$$

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

$$\angle$$ BAD = $$\angle$$ APC (Corresponding angles) ……………….. (i)

And, $$\angle$$ DAC = $$\angle$$ ACP (Alternate interior angles) …….… (ii)

By the new figure, we have;
AP = AC
$$\angle$$ APC = $$\angle$$ ACP ……………………. (iii)

On comparing equations (i), (ii), and (iii), we get,
$$\angle$$ BAD = $$\angle$$ APC

Therefore, AD is the bisector of the angle BAC.
Hence proved.