ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar(ΔADX) = ar(ΔACY)

Given : ABCD is a trapezium with AB || DC and XY || AC. Join YA, XC and XD.
Triangles ACX and ACY have same base AC and are between same parallels AC and XY
So, ar(ΔACX) = ar(ΔACY). (i)
Triangles ACX and ADX have same base AX and are between same parallels AB and DC.
So, ar(ΔACX) = ar(ΔADX). (ii)
From (i) and (ii), ar(ΔADX) = ar(ΔACY)