$A, B, C, D$ - consecutive vertices of a parallelogram. Points $E, F, P, H$ lie on sides $A B$, $B C, C D, A D$ respectively. Segment $A E$ is $\frac{1}{3}$ of side $A B$, segment $B F$ is $\frac{1}{3}$ of side $B C$, and points $P$ and $H$ bisect the sides they lie on. Find the ratio of the area of quadrilateral $E F P H$ to the area of parallelogram $A B C D$.
$S_{\triangle \mathrm{BEF}}=\frac{B E}{A B} \cdot \frac{B F}{B C} S_{\triangle \mathrm{ABC}}$

## Solution

Let $S_{\mathrm{ABCD}}=S$. Then

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} S, S_{\triangle \mathrm{BEF}}=\frac{B E}{A B} \cdot \frac{B F}{B C} S_{\triangle \mathrm{ABC}}=\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{2} S=\frac{1}{9} S
$$

Similarly,

$$
S_{\Delta \mathrm{PDH}}=\frac{1}{8} S, S_{\triangle \mathrm{FCP}}=\frac{1}{6} S, S_{\Delta \mathrm{EAH}}=\frac{1}{12} S .
$$

Therefore,

$$
S_{\mathrm{EFPH}}=S-\frac{1}{9} S-\frac{1}{8} S-\frac{1}{6} S-\frac{1}{12} S=\frac{37}{72} S
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-56.jpg?height=323&width=852&top_left_y=1176&top_left_x=612)

## Answer