# Prove that intersection of diagonals in trapezium divide parallel segment to equal segments

I've got exercise to do as en exercise to my school leaving exam and I have no idea how to prove it:

Diagonals of trapezium intersect in point $S$. Through point $S$ the segment was given that is parallel to the bases and intersect the legs in points $E$ and $F$. Prove that $|ES| = |SF|$.

I'm sorry for my poor translation but I'm not good at math english yet.

Let the trapezium be $ABCD$ and $AB||CD$.As triangles $BSF$ and $BDC$ are similar, $\frac{SF}{DC}=\frac{BS}{BD}\dots\boxed{1}$ and similarly, $\frac{ES}{DC}=\frac{AS}{AC}\dots\boxed{2}$.Now notice that $\triangle ABS\sim\triangle CDS$ and try to finish it off yourself.

Sol:

Given: ABCD is a trapezium.

To Prove: DE/EB = CE /EA

Construction: Draw EF || BA || CD, intersecting AD in F.

Proof:

FE || AB (Given)

DE/EB = DF/FA [According to basic proportionality theorem] --------- (1)

FE || DC (Given)

CE/EA = DF/FA [According to basic proportionality theorem] --------- (2)

From (1) and (2), we get

DE/EB = CE/EA

So, diagonals of a trapezium divide each other proportionally.