So, ar(△ACX)=ar(△ACY) ......(i)
Triangles ACX and ADX have same base AX and are between same parallels AB and DC.
So, ar(△ACX)=ar(△ADX) ......(ii)
From (i) and (ii),
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).
[Hint: Join CX.]