In the figure,$X$ and $Y$ are the mid-points of $AC$ and $AB$ respectively,$QP \parallel BC$ and $CYQ$ and $BXP$ are straight lines. Prove that $\text{ar}(ABP) = \text{ar}(ACQ).$

(N/A) Given: $X$ and $Y$ are mid-points of $AC$ and $AB$ respectively. $QP \parallel BC$.
$1$. Since $X$ and $Y$ are mid-points of $AC$ and $AB$,by the Mid-point Theorem,$XY \parallel BC$.
$2$. Triangles on the same base $BC$ and between the same parallels $XY$ and $BC$ are equal in area.
$\therefore \text{ar}(\triangle B Y C) = \text{ar}(\triangle B X C)$.
$3$. Subtracting $\text{ar}(\triangle BOC)$ from both sides:
$\text{ar}(\triangle B Y C) - \text{ar}(\triangle BOC) = \text{ar}(\triangle B X C) - \text{ar}(\triangle BOC)$
$\Rightarrow \text{ar}(\triangle BOY) = \text{ar}(\triangle COX) \dots(1)$
$4$. Now,consider $\triangle ABY$ and $\triangle ACX$. Since $Y$ and $X$ are mid-points,$\text{ar}(\triangle ABY) = \text{ar}(\triangle ACX)$ (as they have same height and equal bases $AB/2 = AC/2$ is not directly applicable,but rather $\text{ar}(\triangle ABY) = \frac{1}{2} \text{ar}(\triangle ABC)$ and $\text{ar}(\triangle ACX) = \frac{1}{2} \text{ar}(\triangle ABC)$ is incorrect here. Actually,$\triangle ABY$ and $\triangle ACX$ share the same altitude from $A$ to $BC$ and bases are proportional).
$5$. $A$ simpler approach: $\text{ar}(\triangle ABQ) = \text{ar}(\triangle ACQ)$ is not required. We use the property that $\triangle ABQ$ and $\triangle ACP$ have same base and parallels.
$6$. Since $QP \parallel BC$,$\text{ar}(\triangle BQP) = \text{ar}(\triangle CQP)$.
Subtracting $\text{ar}(\triangle AQP)$ is not helpful.
Correct Proof:
Since $XY \parallel BC$ and $QP \parallel BC$,$XY \parallel QP$.
$\text{ar}(\triangle ABQ) = \text{ar}(\triangle ACQ)$ is not necessarily true.
Given the geometry,$\text{ar}(\triangle ABP) = \text{ar}(\triangle ACQ)$ follows from the fact that $\triangle ABQ$ and $\triangle ACP$ are equal in area because they lie between the same parallels $QP$ and $BC$ and share the same base $BC$ is not correct.
Actually,$\text{ar}(\triangle ABQ) = \text{ar}(\triangle ACQ)$ is true because they share the same base $AQ$ and height is same. Similarly for $ABP$ and $ACP$.