Question

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

Answer 1

Given:- Let $ABC$ be a triangle where $DE\parallel BC$ and $D$ is the mid point of $AB$.

To prove:- $E$ is the mid-point of $AC$.

Proof:-

In $△ABC$

$DE\parallel BC$

We know that if a line drawn parallel to one side of a triangle intersects the other two sides in distinct points, then it divides the other side in same ratio.

$\therefore \frac{AD}{DB}=\frac{AE}{EC}.....\left(1\right)$

Since $D$ is the mid point of $AB$

$\therefore AD=DB$

From ${eq}^n\left(1\right)$, we have

$AE=EC$

Hence $E$ is the mid-point of $AC$.

Hence proved.