4. On the sides $B C$ and $A D$ of the convex quadrilateral $A B C D$, their midpoints - points $M$ and $N$ respectively - are marked. Segments $M N$ and $A C$ intersect at point $O$, and $M O=O N$. It is known that the area of triangle $A B C$ is 2017. Find the area of quadrilateral $A B C D$.
Answer: 4034.

## Solution:

Let $A C$ and $M N$ intersect at point $O$ (see the first figure on the right), $S_{\triangle A B C}=2017=S$.

We will prove that $S_{\triangle A D C}=S_{\triangle A B C}$, then $S_{A B C D}=2 S$.

There are various ways to reason.

![](https://cdn.mathpix.com/cropped/2024_05_06_501086c9648264b65f19g-4.jpg?height=343&width=711&top_left_y=248&top_left_x=1232)

First method.

Draw $A M$ and $C N$ (see the first figure). Since $A O$ is the median of triangle $A M N$, then $S_{\triangle А o N}=S_{\triangle А o м}$. Similarly, $S_{\triangle C O N}=S_{\triangle С o м}$.

Therefore, $S_{\triangle A N C}=S_{\triangle A M C}$. Since $A M$ and $C N$ are medians of triangles $A B C$ and $A D C$ respectively, then $S_{\triangle A D C}=2 S_{\triangle A N C}=2 S_{\triangle A M C}=S_{\triangle A B C}=S$.

Second method.

Let $K$ and $L$ be the midpoints of sides $A B$ and $C D$, then $K M L N$ is a parallelogram, so segment $K L$ contains point $O$ (see the second figure).

Since $K L$ is the midline of triangle $A B C$, then $S_{\triangle K O M}=\frac{1}{4} S_{\triangle A B C}=\frac{1}{4} S$.

![](https://cdn.mathpix.com/cropped/2024_05_06_501086c9648264b65f19g-4.jpg?height=351&width=702&top_left_y=984&top_left_x=1245)

Similarly, $S_{\triangle N O L}=\frac{1}{4} S_{\triangle A D C}$. Since $S_{\triangle N O L}=S_{\triangle K O M}$, then $S_{\triangle A D C}=S_{\triangle A B C}=S$.

Note that $K M L N$ is the Varignon parallelogram, the area of which is half the area of $A B C D$.

Third method.

We use the fact that in any convex quadrilateral $A B C D$, the midpoints $M$ and $N$ of opposite sides and the midpoints $P$ and $Q$ of the diagonals are the vertices of a parallelogram (see the third figure).

Therefore, the midpoint $O$ of segment $M N$ is also the midpoint of segment $P Q$.

Since in our case point $O$ lies on diagonal $A C$, then the midpoint $Q$ of diagonal $B D$ also lies on $A C$ (see the fourth

![](https://cdn.mathpix.com/cropped/2024_05_06_501086c9648264b65f19g-4.jpg?height=432&width=471&top_left_y=1600&top_left_x=1478)
figure).

$$
\begin{aligned}
& \text { Then } S_{\triangle A D Q}=S_{\triangle A B Q} \text { and } S_{\triangle C D Q}=S_{\triangle C B Q} \cdot \\
& \text { Therefore, } S_{\triangle A D C}=S_{\triangle A B C}=S .
\end{aligned}
$$

## Criteria:

A correct answer without a correct justification is not scored. For proving the equality of areas of separate elements of the triangle without further correct progressions (without a complete solution), no more than 2 points in total should be given.