Therefore, ax = xc or AEAB = CDED.

Therefore, a
x = x
c or AE
AB = CD
ED. Since ∠BAE and ∠CDE are base angles of the
isosceles trapezoid, then triangle BAE is similar to triangle EDC.
Next we consider triangles CQE and CQD where Q is the intersection of a
perpendicular dropped from C to base AD. If CQ = h, then QD = e−b
2 = a+c−b
2
so that EQ = ED − QD = c−a+b
2 . By the Pythagorean Theorem for triangles
CQE and CQD respectively, we have z2 = h2 +