1.29. Through points $R$ and $E$, belonging to sides $A B$ and $A D$ of parallelogram $A B C D$ and such that $A R=\frac{2}{3} A B, A E=\frac{1}{3} A D$, a line is drawn. Find the ratio of the area of the parallelogram to the area of the resulting triangle.
1.29. Let $h$ be the height of parallelogram $ABCD$, and $h_{1}$ be the height of triangle $ARE$ (Fig. 1.25). Then $S_{ABCD} = AD \cdot h$, $S_{\triangle ARE} = \frac{1}{2} AE \cdot h_{1}$. But $\frac{h}{h_{1}} = \frac{AB}{AR} = \frac{3}{2}$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-426.jpg?height=354&width=523&top_left_y=1504&top_left_x=627)

Fig. 1.25

$$
\frac{S_{ABCD}}{S_{\triangle ARE}} = \frac{AD \cdot h}{\frac{1}{2} AE \cdot h_{1}} = \frac{AD \cdot h}{\frac{1}{2} \cdot \frac{1}{3} AD \cdot \frac{2}{3} h} = 9
$$

Answer: $9:1$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-427.jpg?height=415&width=646&top_left_y=396&top_left_x=46)

Fig. 1.26