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 XEF similar XDC and 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