Question 846102
Suppose that in trapezoid *[tex \large ABCD], *[tex \large AB || CD] and *[tex \large AC = BD]. Denote by E the intersection of AC and BD.


Because triangles ABE and CDE are similar by angle-angle-angle, we can write DE = x, CE = y, then AE = ky and BE = kx. Since AC = BD, we have ky + y = kx + x --> y(k+1) = x(k+1), which implies x = y. Therefore DE = CE and AE = BE.


It follows that triangles ABE and CDE are isosceles, so *[tex \large \angle CAB = \angle DBA], and by SAS, triangles ABC and BAD are congruent, so AD = BC. The trapezoid is isosceles.