The midline $K L$ of the equilateral triangle $A B C$ is also the midline of triangle $D E F$, in which vertex $D$ lies on the segment $A C$, and vertex $F$ lies on the extension of side $A C$ beyond point $C$. The area of quadrilateral $D K L C$ is $\frac{3}{8}$ of the area of triangle $D E F$. Find the angle $E D F$.

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Let $P$ be the projection of point $K$ onto $A C$. Prove that $D P=\frac{1}{2} A C$.

## Solution

Notice that points $B$ and $E$ lie on the same side of line $A C$, and triangles $A B C$ and $D E F$ are equal in area, as they have equal bases ($A C=D F$) and heights. Therefore,

$$
\begin{aligned}
S_{\triangle \mathrm{DKA}} & =S_{\mathrm{CLKA}}-S_{\mathrm{DKLC}}=\frac{3}{4} S_{\Delta \mathrm{ABC}}-\frac{3}{8} S_{\Delta \mathrm{EDF}}= \\
& =\frac{3}{4} S_{\Delta \mathrm{ABC}}-\frac{3}{8} S_{\Delta \mathrm{ABC}}=\frac{3}{8} S_{\Delta \mathrm{ABC}}
\end{aligned}
$$

Let $P$ be the projection of point $K$ onto $A C$. Then

$$
S_{\triangle \mathrm{DKP}}=S_{\triangle \mathrm{DKA}}-S_{\triangle \mathrm{PKA}}=\frac{3}{8} S_{\triangle \mathrm{ABC}}-\frac{1}{8} S_{\triangle \mathrm{ABC}}=\frac{1}{4} S_{\triangle \mathrm{ABC}} .
$$

Let $A C=a$. Then

$$
S_{\triangle \mathrm{ABC}}=\frac{a^{2} \sqrt{3}}{4}, K P=\frac{a \sqrt{3}}{4}, \frac{1}{2} D P \cdot K P=\frac{1}{4} S_{\triangle \mathrm{ABC}}, \text { or } \frac{1}{2} D P \cdot \frac{a \sqrt{3}}{4}=\frac{1}{4} \cdot \frac{a^{2} \sqrt{3}}{4} .
$$

From this, we find that $D P=\frac{a}{2}$. Therefore,

$$
\operatorname{tg} \angle K D A=\frac{K P}{P D}=\frac{\frac{a \sqrt{3}}{4}}{\frac{a}{2}}=\frac{\sqrt{3}}{2}, \angle E D F=180^{\circ}-\angle K D A=180^{\circ}-\operatorname{arctg} \frac{\sqrt{3}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e3b1ddf947c7dd2094eeg-09.jpg?height=412&width=994&top_left_y=0&top_left_x=535)

## Answer

$\pi-\operatorname{arctg} \frac{\sqrt{3}}{2}$.

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