In $\Delta ABC,$ $P$ and $Q$ are the points of trisection of $BC$ (i.e.,points dividing $BC$ into three equal parts). Prove that,$\operatorname{ar}(ABP) = \operatorname{ar}(APQ) = \operatorname{ar}(AQC) = \frac{1}{3} \operatorname{ar}(ABC).$

(N/A) $P$ and $Q$ are the points of trisection of $BC$.
Therefore,$BP = PQ = QC$.
Since the triangles $\Delta ABP, \Delta APQ,$ and $\Delta AQC$ have equal bases $(BP = PQ = QC)$ and share the same vertex $A$,they have the same altitude with respect to these bases.
We know that the area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Since the bases are equal and the height is common,the areas of these three triangles are equal.
$\operatorname{ar}(ABP) = \operatorname{ar}(APQ) = \operatorname{ar}(AQC)$.
Also,the sum of the areas of these three triangles equals the area of $\Delta ABC$:
$\operatorname{ar}(ABP) + \operatorname{ar}(APQ) + \operatorname{ar}(AQC) = \operatorname{ar}(ABC)$.
Substituting the equal areas,we get:
$3 \times \operatorname{ar}(ABP) = \operatorname{ar}(ABC)$.
Therefore,$\operatorname{ar}(ABP) = \operatorname{ar}(APQ) = \operatorname{ar}(AQC) = \frac{1}{3} \operatorname{ar}(ABC).$