Question

Using Theorem $6.1,$ prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Theorem 6.1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Answer 1

Given:

In $△ABC,D$ is midpoint of $AB$ and $DE$ is parallel to $BC.$
$\therefore$ $AD=DB$

To prove:

$AE=EC$

Proof:

Since, $DE\parallel BC$
$\therefore$ By Basic Proportionality Theorem,
$\frac{AD}{DB}=\frac{AE}{EC}$

Since, $AD=DB$
$\therefore$ $\frac{AE}{EC}=1$
$\therefore$ $AE=EC$