$P$ and $Q$ are respectively the mid-points of sides $AB$ and $BC$ of a triangle $ABC$ and $R$ is the mid-point of $AP$. Show that $\text{ar} (PBQ) = \text{ar} (ARC)$.

(N/A) Given: $P$ is the mid-point of $AB$,$Q$ is the mid-point of $BC$,and $R$ is the mid-point of $AP$.
$1$. In $\Delta ABC$,$AQ$ is a median (since $Q$ is the mid-point of $BC$). Therefore,$\text{ar}(\Delta ABQ) = \frac{1}{2} \text{ar}(\Delta ABC)$.
$2$. In $\Delta ABQ$,$PQ$ is a median (since $P$ is the mid-point of $AB$). Therefore,$\text{ar}(\Delta PBQ) = \frac{1}{2} \text{ar}(\Delta ABQ) = \frac{1}{2} \times (\frac{1}{2} \text{ar}(\Delta ABC)) = \frac{1}{4} \text{ar}(\Delta ABC)$.
$3$. In $\Delta ABC$,$AC$ is a side. Consider $\Delta ARC$. Since $R$ is the mid-point of $AP$ and $P$ is the mid-point of $AB$,$AR = RP = PB = \frac{1}{4} AB$.
$4$. Alternatively,using the property of medians: In $\Delta APC$,$CR$ is a median. Thus,$\text{ar}(\Delta ARC) = \frac{1}{2} \text{ar}(\Delta APC)$.
$5$. Since $P$ is the mid-point of $AB$,$\text{ar}(\Delta APC) = \frac{1}{2} \text{ar}(\Delta ABC)$.
$6$. Therefore,$\text{ar}(\Delta ARC) = \frac{1}{2} \times (\frac{1}{2} \text{ar}(\Delta ABC)) = \frac{1}{4} \text{ar}(\Delta ABC)$.
Thus,$\text{ar}(\Delta PBQ) = \text{ar}(\Delta ARC) = \frac{1}{4} \text{ar}(\Delta ABC)$.