### In the given figure, $ABCD$ is a parallelogram whose diagonals $AC$ and $BD$ intersect at $O$. A line segment through $O$ meets $AB$ at $P$ and $DC$ at $Q$. Prove that $ar(\square APQD)$ = $\dfrac { 1 } { 2 } ar(||gm\space ABCD)$ D C Q P A B O

1. We know that diagonal $AC$ of $||gm \space ABCD$ divides it into two triangles of equal area. \begin{aligned} \therefore ar(\Delta ACD) = \dfrac { 1 } { 2 } ar(||gm \space ABCD) &&\ldots \text{(i)} \end{aligned}
2. Now, In $\Delta OAP$ and $\Delta OCQ$, we have: \begin{aligned} &OA = OC &&[\text{Diagonals of a ||gm bisect each other}] \\ &\angle AOP = \angle COQ &&[\text{Vertically opposite angles}]\\ &\angle PAO = \angle QCO &&[\text{Alternate interior angles}]\\ \therefore &\Delta OAP \cong \Delta OCQ \end{aligned}
3. We know that if two triangles are congurrent then their respective areas are equal. \begin{aligned} \therefore& \space ar(\Delta OAP) = ar(\Delta OCQ) \\ \implies& ar(\Delta OAP) + ar( \square AOQD) = ar(\Delta OCQ) + ar(\square AOQD) \\ \implies& ar(\square APQD) = ar(\Delta ACD) \\ &\space\space\space\space\space\space\space\space\space\space\space\space\space \space\space\space\space\space\space\space\space\space = \dfrac { 1 } { 2 } ar(||gm\space ABCD) &&[\text{Using eq (i)}] \\ \therefore&\space ar(\square APQD) = \dfrac { 1 } { 2 } ar(||gm\space ABCD) \end{aligned} 