Question

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

Answer 1

Step 1: Note the given data

Let $ABCD$ be a trapezium with $AB\parallel DC$

$AD$ and $BC$ are non-parallel sides.

Let, $EF$ be a line parallel to $AB,DC$.

Join $AC$ such that $△ACD$ and $△ACB$ are triangles and $AC$ meets $EF$ at $G$.

Step 2: Apply Thales theorem on $∆ACD$ and $∆ACB$

Basic Proportionality Theorem: In a triangle, a line drawn parallel to one side to intersect the other sides distinct points divides two sides in same ratio.

In $∆ACD$

Here, $EG\parallel DC$. According to Basic Proportionality Theorem

$\frac{AE}{ED}=\frac{AG}{GC}$……………..(i)

In $∆ACB$

Here, $GF\parallel AB$. According to Basic Proportionality Theorem

$\begin{array}{l}\frac{FC}{BF}=\frac{GC}{AG}\\ \Rightarrow \frac{BF}{FC}=\frac{AG}{GC}......................\left(ii\right)\end{array}$

Equating equation (i) and equation (ii)

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

Hence proved that a line drawn parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.