Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side.(Recall that you have proved it in Class IX).

Given, in \(\triangle ABC\), D is the midpoint of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.

We have to prove that E is the mid point of AC.

Since, D is the mid-point of AB.
AD=DB
=>\(\frac{AD}{DB}\) = 1 …………………………. (i)

In \(\triangle ABC\), DE || BC,

By using Basic Proportionality Theorem,

Therefore, \(\frac{AD}{DB} = \frac{AE}{EC}\)

From equation (i), we can write,
=> 1 = \(\frac{AE}{EC}\)
=> AE = EC

Hence, proved, E is the midpoint of AC.