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

The segments connecting the midpoints of opposite sides of a convex quadrilateral $ABCD$ are perpendicular, $AC=4, \angle CAB + \angle DBA = 75^{\circ}$. Find the area of quadrilateral $ABCD$ and compare it with the number $2 \sqrt{15}$.
A quadrilateral with vertices at the midpoints of the sides of quadrilateral $A B C D$ is a rhombus.

## Solution

Let $K, L, M$ and $N$ be the midpoints of the sides $A B, B C, C D$ and $A D$ of the given quadrilateral, respectively. Since $K N$ is the midline of triangle $A B D$, we have $K N \| B D$ and $K N=\frac{1}{2} B D$. Similarly, $L M \| B D$ and $L M=\frac{1}{2} B D$.

Therefore, quadrilateral $K L M N$ is a parallelogram. Since its diagonals $K M$ and $L N$ are perpendicular, it is a rhombus. The diagonals of quadrilateral $A B C D$ are twice the length of the sides of the rhombus, so $B D=A C=4$. The acute angle between the diagonals is $75^{\circ}$ (by the theorem on the exterior angle of a triangle). Therefore,

$$
\begin{gathered}
S_{\mathrm{ABCD}}=\frac{1}{2} \cdot A C \cdot B D \cdot \sin 75^{\circ}=\frac{1}{2} \cdot 4 \cdot 4 \cdot \frac{\sqrt{6}+\sqrt{2}}{4}=2 \sqrt{2}(\sqrt{3}+1) \\
2 \sqrt{2}(\sqrt{3}+1)<2 \sqrt{15} \Leftarrow(\sqrt{6}+\sqrt{2})^{2}<(\sqrt{15})^{2} \Leftarrow \\
8+4 \sqrt{3}<15 \Leftarrow 4 \sqrt{3}<7 \Leftarrow 48<49
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-33.jpg?height=766&width=824&top_left_y=0&top_left_x=617)

## Answer

$2 \sqrt{2}(\sqrt{3}+1)<2 \sqrt{15}$

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