In the figure,$ABCD$ is a parallelogram and $BC$ is produced to a point $Q$ such that $AD = CQ$. If $AQ$ intersects $DC$ at $P$,show that $\operatorname{ar}(BPC) = \operatorname{ar}(DPQ)$. [Hint: Join $AC$.]

(N/A) We have a parallelogram $ABCD$ and $AD = CQ$.
Let us join $AC$. We know that triangles on the same base and between the same parallels are equal in area.
Since $\Delta ADC$ and $\Delta ABC$ are on the same base $AC$ and between the same parallels $AB$ and $DC$,we have $\operatorname{ar}(\Delta ADC) = \operatorname{ar}(\Delta ABC)$.
Also,since $AD \parallel BQ$ and $AD = CQ$,we consider $\Delta ADC$ and $\Delta QDC$. However,a more direct approach is to use the property that triangles on the same base and between the same parallels are equal in area.
Consider $\Delta ADC$ and $\Delta ACQ$. Since $AD \parallel BQ$ (as $AD \parallel BC$),$\Delta ADC$ and $\Delta ACQ$ are between the same parallels $AD$ and $BQ$. But they do not share the same base.
Let us use the property: $\operatorname{ar}(\Delta ADC) = \operatorname{ar}(\Delta ABC)$.
Adding $\operatorname{ar}(\Delta APC)$ to both sides is not helpful. Instead,consider $\Delta ADQ$ and $\Delta ACQ$. Since $AD \parallel QC$,$\operatorname{ar}(\Delta ADQ) = \operatorname{ar}(\Delta ADC)$.
Actually,the standard proof is:
$1$. $\operatorname{ar}(\Delta ADC) = \operatorname{ar}(\Delta ABC)$ (Triangles on same base $AC$ and between parallels $AB \parallel DC$)
$2$. Since $AD \parallel BQ$,$\operatorname{ar}(\Delta ADQ) = \operatorname{ar}(\Delta ACQ)$ (Triangles on same base $AQ$ and between parallels $AD \parallel BQ$ is not correct here).
Correct approach:
Since $AD \parallel BQ$,$\operatorname{ar}(\Delta ADQ) = \operatorname{ar}(\Delta ACQ)$ is false.
Correct logic: $\operatorname{ar}(\Delta ADC) = \operatorname{ar}(\Delta ACQ)$ is false.
Let's use: $\operatorname{ar}(\Delta ADQ) = \operatorname{ar}(\Delta ACQ)$ is not true.
Actually,$\operatorname{ar}(\Delta ADQ) = \operatorname{ar}(\Delta ACQ)$ is true if $AD \parallel QC$. Yes,$AD \parallel BC$,so $AD \parallel QC$. Thus,$\operatorname{ar}(\Delta ADQ) = \operatorname{ar}(\Delta ACQ)$.
Subtracting $\operatorname{ar}(\Delta APD)$ from both sides:
$\operatorname{ar}(\Delta ADQ) - \operatorname{ar}(\Delta APD) = \operatorname{ar}(\Delta ACQ) - \operatorname{ar}(\Delta APD)$
$\operatorname{ar}(\Delta DPQ) = \operatorname{ar}(\Delta APC)$
Since $\operatorname{ar}(\Delta APC) = \operatorname{ar}(\Delta BPC)$ (triangles on same base $PC$ and between parallels $AB \parallel DC$),
Therefore,$\operatorname{ar}(\Delta DPQ) = \operatorname{ar}(\Delta BPC)$.