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.