Question

In the following figure, CE is drawn parallel to diagonal DB of the quadrilateral ABCD which meets AB produced at point E. Prove that $\Delta$ ADE and quadrilateral ABCD are equal in area.

Answer 1

$ABCD$ is a quadrilateral where $BE\parallel AC.$

Since $△BAC$ and $△EAC$ lie on the same base $AC$ and between the same parallel $AC$ and $BE$

$\therefore ar\left(BAC\right)=ar\left(EAC\right)$

Triangle with same base and between the same parallels are equal in area.

Adding $ar\left(ADC\right)$ on both sides

$\Rightarrow ar\left(BAC\right)+ar\left(ADC\right)=ar\left(EAC\right)+ar\left(ADC\right)$

$\Rightarrow ar\left(ABCD\right)=ar\left(ADE\right)$

Hence, the answer is solved.