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

Construction: Join CX.
Triangles ADX and ACX are on the same base AX and between same parallels AB ∥ DC.
Hence, ar(ADX) = ar(ACX) …(1)
[∵ Triangles on the same base (or equal bases) and between same parallels are equal in area.]
Similarly, triangles ACY and ACX are on the same base AC and between same parallels AC ∥ XY.
Hence, ar(ACY) = ar(ACX) …(2)
From the equation (1) and (2), ar(ADX) = ar(ACY).