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.]

It can be observed that, \(\triangle{ADX}\) and \(\triangle{ACX}\) lie on the same base AX and are between the same parallels AB and DC.

\(\therefore \) Area (\(\triangle{ADX}\)) = Area (\(\triangle{ACX}\)) ...(i)

Also, \(\triangle{ADY}\) and \(\triangle{ACX}\) lie on the same base AC and are between the same parallels AC and XY.

\(\therefore \) Area (\(\triangle{ACY}\)) = Arca (\(\triangle{ACX}\)) ...(ii)

From Equations (i) and (ii), we get,

Area (\(\triangle{ADX}\)) = Area (\(\triangle{ACY}\))

Hence, proved.