Question

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

Answer 1

For $△ACX$ and $△ACY$ lie on the same base AC and are between parallel lines AC and XY.

$ar\left(ACX\right)=ar\left(ACY\right).....\left(1\right)$ [Traingle with same base and between the same parallels are equal in area]

Also, for $△ACX$ and $△ADX$ lie on the same base AX and between parallel lines AX and DC [As $AB||DC$, parts of parallel lines are parallel]

$ar\left(ACX\right)=ar\left(ADX\right)......\left(2\right)$ [Traingle with same base and between the same parallels are equal in area]

From (1) and (2)

$ar\left(ADX\right)=ar\left(ACY\right)$

Hence proved.