Let ABCD be a trapezoid, with AB parallel to CD.
Draw the diagonals AC and BD and let them cross at O.
Draw the line through O parallel to AB, and let it cross AD and BC
at F and G respectively.

Let a = length of AB, b = length of CD and c = length of FG.
We are asked to express c in terms of a and b.

Draw the line KL through O perpendicular to AB.

Triangle OKB is similar to triangle OLD, so OK/OL = OB/OD,
and triangle OBA is similar to triangle ODC, so OB/OD = AB/CD = a/b.
Hence OK/OL = a/b, and so OK = KL * a/(a+b) and OL = KL * b/(a+b).

The area of ABCD = (a + b)*KL/2, and this must equal the area of
ABFG plus the area of FGDC.
But the area of ABFG = (a + c)*OK/2 = a(a+c)*KL/2(a+b), and the
area of FGDC = (c + b)*OL/2 = b(b+c)*KL/2(a+b).
Hence:

(a+b)/2 = a(a+c)/2(a+b) + b(b+c)/2(a+b)

a^{2} + 2ab + b^{2} = a^{2} + (a+b)c + b^{2}

c = 2ab/(a+b)

We can write this as 1/c = (1/a + 1/b)/2, showing that c is in fact
the harmonic mean of a and b!