10.157. Through points $R$ and $E$, belonging to sides $A B$ and $A D$ of parallelogram $A B C D$, and such that $A R=(2 / 3) A B, A E=(1 / 3) A D$, a line is drawn. Find the ratio of the area of the parallelogram to the area of the resulting triangle.
## Solution.

Let $h$ be the height of parallelogram $ABCD$, and $h_{1}$ be the height of triangle $ARE$ (Fig. 10.134). Then $S_{ABCD} = AD \cdot h$, and $S_{\triangle ARE} = \frac{1}{2} AE \cdot h_{1}$. But $\frac{h}{h_{1}} = \frac{AB}{AR} = \frac{3}{2}$. Therefore, $\frac{S_{ABCD}}{S_{\triangle ARE}} = \frac{AD \cdot h}{\frac{1}{2} AE \cdot h_{1}} = \frac{AD \cdot h}{\frac{1}{2} \cdot \frac{1}{3} AD \cdot \frac{2}{3} h} = 9$.

Answer: 9.