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,

AD || PC
\(\angle\) 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.