10.1. The area of the quadrilateral formed by the midpoints of the bases and diagonals of a trapezoid is four times smaller than the area of the trapezoid itself. Find the ratio of the lengths of the bases of the trapezoid.
Answer: $3: 1$.

Solution. Let the vertices of the trapezoid be $A, B, C, D$, the midpoints of the bases $A D$ and $B C$ be $K$ and $M$, and the midpoints of the lateral sides $A B$ and $C D$ be $P$ and $Q$ respectively. Let the lengths of $A D$ and $B C$ be $a$ and $b$, with $a > b$. By Thales' theorem, the midline $P Q$ intersects the diagonals $A C$ and $B D$ at their midpoints $L$ and $N$. Then $L Q$ and $N Q$ are the midlines of triangles $A C D$ and $B C D$, so $L N = L Q - N Q = \frac{a}{2} - \frac{b}{2} = \frac{a - b}{2}$. If the height of the trapezoid is $h$, then the perpendiculars from points $K$ and $M$ to the midline $P Q$ are $\frac{h}{2}$, and the area of the quadrilateral $K L M N$, formed by the midpoints of the bases and diagonals of the trapezoid, is $\frac{1}{2} \cdot h \cdot \frac{a - b}{2} = h \frac{(a - b)}{4}$. The area of the trapezoid is $h \frac{(a + b)}{2}$, from which $\frac{a - b}{2(a + b)} = \frac{1}{4}$ and $a = 3 \mathrm{~b}$.

Grading criteria. Proof that the length of $L N$ is half the difference of the bases: 2 points.

Calculation of the area of the quadrilateral $K L M N$: 3 points.

Completion of the calculation: 2 points.