3. In a convex quadrilateral $A B C D$, the midpoints of consecutive sides are marked: $M, N, K, L$. Find the area of quadrilateral $M N K L$, if $|A C|=|B D|=2 a,|M K|+|N L|=2 b$.


Solution. Since $M N, N K, K L$ and $L M$ are the midlines of triangles $A B C, B C D, C D A$ and $D A B$ respectively, then $M N K L$ is a rhombus with side length $a$. Let $O$ be the point of intersection of its diagonals and consider triangle $M O N$. Since the diagonals of a rhombus are perpendicular to each other, $\triangle M O N$ is a right triangle. Let the lengths of its legs be $x, y$, then the area of $\triangle M O N$ is $\frac{1}{2} x y$, hence the area of quadrilateral $M N K L$ is $2 x y$. On the other hand, by the problem's condition, $x^{2}+y^{2}=a^{2}$, $x+y=b$. From this, it is easy to find that $2 x y=b^{2}-a^{2}$, i.e., the area of quadrilateral $M N K L$ is $b^{2}-a^{2}$.

Grading criteria. It is established that $M N K L$ is a rhombus, - 1 point. All relations that allow finding the area of $M N K L$ purely algebraically (the area is $2 x y$, where $x^{2}+y^{2}=a^{2}, x+y=b$) are written, but the solution is not found - 4 points. Only the answer without progress in the solution 0 points.