Question

Prove that any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.

Answer 1

Let $ABCD$ be a trapezium $DC\parallel AB$ and $EF$ is a line parallel to $DC$ and $AB$.

To prove :

$\frac{AE}{ED}=\frac{BF}{FC}$

Construction :

Join $AC$, meeting $EF$ in $G$.

Proof :

In $△ADC$, we have

$EG\parallel DC$

$\frac{AE}{ED}=\frac{AG}{GC}$ [By Thale's theorem] .....$\left(1\right)$

In $△ABC$, we have

$GF\parallel AB$

$\frac{AG}{GC}=\frac{BF}{FC}$ [By Thales's theorem] ......$\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$, we get,

$\frac{AE}{ED}=\frac{BF}{FC}$

Hence, it is proved that any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.