Trapezoid midpoints of diagonals

ABCD is a trapezoid with AB parallel to DC and AB > DC

Draw line from C through F to meet AB at G

Draw line from E to meet AB at H parallel to CF

AE = EC & BF = FD (Construct)

Consider triangles FGB, FCD

^GBF = ^CDF (Inclusive angles within parallel lines AB, DC)

FB = FD (Construct)

^GFB = ^DFC (Opp. angles of intersecting lines are equal)

So

So

BG = DC

EFGH is a parallelogram (Opp. sides of quad. parallel)

So

HG = EF (Opp. sides of parallelogram are equal)

Consider triangles AEH, ECF

AE = EC (Construct)

AC and GC are transversals of parallel lines AB and DC.

So

HE = GF (Opp. sides of parallelogram are equal)

So

HE = FC

^AEH = ^ECF (Corresponding angles EF parallel to DC)

So

So

EF = AH

But we have already shown that EF = HG

So

EF = 1/2(AG)

AG = AB - BG = AB - DC (Already shown BG = DC)

So

EF = 1/2(AB - DC)