In the figure, D is a point on side BC of triangle ABC such that \(BD\over CD\) = \(AB\over AC\). Prove that AD is the bisector of \(\angle\) BAC.

Solution :

Given : ABC is a triangle and D is point on BC such that \(BD\over CD\) = \(AB\over AC\)

To Prove : AD is the bisector of \(\angle\) BAC.

Construction : Produce line BA to E such that line AE = AC. Join CE.

Proof : In \(\triangle\) AEC, since AE = AC, hence

\(\angle\) AEC = \(\angle\) ACE      (angles opposite to equal sides of triangle are equal)

Now,  \(BD\over CD\) = \(AB\over AC\)     (given)

So,  \(BD\over CD\) = \(AB\over AE\)          (AE = AC, by construction)

\(\therefore\)  By converse of Basic Proportionality theorem(Thales Theorem),

DA || CE

Now, Since CA is a traversal, we have :

\(\angle\) BAD = \(\angle\) AEC      ……..(2)        [corresponding angles]

and  \(\angle\) DAC = \(\angle\) ACE       ……..(3)      (alternate angles)

Also,  \(\angle\) AEC = \(\angle\) ACE           (from 1)

From (2) and (3),

\(\angle\) BAD = \(\angle\) DAC

Thus,  AD bisects \(\angle\) BAC.