# Prove that the line drawn from the mid-point of one side of triangle parallel to another side bisects the third side.

## Solution :

Given : A triangle ABC in which D is the mid point of side AB and the line DE is drawn parallel to BC, meeting AC in E. To Prove :  AE = EC

Proof : In triangle ABC, DE || BC

By basic proportionality theorem,

$$AD\over DB$$ = $$AE\over EC$$                  ………..(1)

But given that D is the mid point of AB,

AB = DB     $$\implies$$   $$AD\over DB$$ = 1             …….(2)

From (1) and (2), we obtain that

$$AE\over EC$$ = 1          $$\implies$$   AE = EC.

Hence, E bisects AC.
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