# 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.

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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.