[ Varignon's Parallelogram ]  $[$ Area of a Quadrilateral ]

In a convex quadrilateral $A B C D$, point $L$ is the midpoint of side $B C$, point $M$ is the midpoint of $A D$, and point $N$ is the midpoint of side $A B$. Find the ratio of the area of triangle $L M N$ to the area of quadrilateral $A B C D$.
Let $K$ be the midpoint of side $CD$. Then $NLKM$ is a parallelogram, and its area is half the area of the given quadrilateral.

## Solution

Let $K$ be the midpoint of side $CD$, $\alpha$ be the angle between the diagonals $AC$ and $BD$ of quadrilateral $ABCD$. Since $NL$ and $MK$ are the midlines of triangles $ABC$ and $ADC$, $NLKM$ is a parallelogram with angle $\alpha$ between adjacent sides.

Since

\[
\begin{gathered}
S_{\mathrm{ABCD}}=\frac{1}{2} AC \cdot BD \sin \alpha \\
S_{\mathrm{NLKM}}=NL \cdot MN \sin \alpha=\frac{1}{4} AC \cdot BD \sin \alpha
\end{gathered}
\]

then

\[
S_{\mathrm{NLKM}}=\frac{1}{2} S_{\mathrm{ABCD}}, S_{\Delta \mathrm{NLM}}=\frac{1}{2} S_{\mathrm{NLKM}}
\]

Therefore, $S_{\Delta \mathrm{LMN}}=\frac{1}{4} S_{\mathrm{ABCD}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-47.jpg?height=474&width=761&top_left_y=2007&top_left_x=657)

## Answer

$\frac{1}{4}$

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