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[SOLVED with many thanks] Trapezium Proof

ABCD is a trapezium, with AB||CD. A is joined to C and B is joined to D. Point E is the midpoint of BD and point F is the midpoint of AC. E is joined to F.

Show that the line EF is half the distance of the difference between DC and AB.

Not sure if this is sound but was how I started:

If E and and F are midpoints of BD and AC respectively, and AB||CD then EF must be parallel to AB and CD (family of parallel lines cut off intercepts in equal ratios)

then i thought that I could use similar triangles. Letting diagonals intersect at X and after proving $\displaystyle \Delta$XEF similar $\displaystyle \Delta$XDC and $\displaystyle \Delta$AXB with sides in equal ratios, but can't figure out how to get it structured right or proceed from there.

Been at it a while now and it perhaps has become a issue of can't see the forest for the trees, but am defnintely hittng a brick wall.

any help greatly appreciated