# 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.