[ Remarkable property of a trapezoid ] [ Transferring sides, diagonals, etc. ]

In quadrilateral $A B C D$, diagonals $A C$ and $B D$ intersect at point $K$. Points $L$ and $M$ are the midpoints of sides $B C$ and $A D$, respectively. Segment $L M$ contains point $K$. Quadrilateral $A B C D$ is such that a circle can be inscribed in it. Find the radius of this circle if $A B=3, A C=\sqrt{13}$, and $L K: K M=1: 3$.
Prove that $A D \| B C$. Denote $B C=x$, express the sides of the trapezoid $A B C D$ in terms of $x$, and use the cosine theorem to form an equation in terms of $x$.

## Solution

We will prove that $B C \| A D$. Suppose this is not the case, and draw a line through point $B$ parallel to $A D$

(figure 1). It is known that the intersection point of the diagonals and the midpoints of the bases of any trapezoid lie on the same line. Therefore, the line $M K$ intersects the segment $B P$ at its midpoint $L_{1}$. Then the segment $L L_{1}$ is the midline of triangle $C B P$. Thus, $L L_{1} \| A C$ and $L M \| A C$, but by the condition, $L M$ and $A C$ intersect at point $K$. A contradiction is obtained. Therefore, the quadrilateral $A B C D$ is a trapezoid with bases $B C$ and $A D$.

From the similarity of triangles $B L K$ and $D M K$ (figure 2), it follows that $B L: D M=K L: K M=1: 3$, and since points $L$ and $M$ are the midpoints of the bases $B C$ and $A D$, then $B C: A D=1: 3$.

Let $B C=x$. Then $A D=3 x$. Since a circle can be inscribed in the quadrilateral $A B C D$, the sums of its opposite sides are equal, i.e., $B C+A D=A B+C D$. Therefore, $C D=B C+A D-A B=x+3 x-3=4 x-3$. Draw a line through vertex $C$ parallel to side $A B$. Let this line intersect the base $A D$ at point $Q$. From the cosine theorem for triangles $A D C$ and $Q D C$, we get

$$
\begin{gathered}
\cos \angle A D C=\frac{D A^{2}+D C^{2}-A C^{2}}{2 \cdot D A \cdot D C}=\frac{9 x^{2}+(4 x-3)^{2}-13}{2 \cdot 3 x(4 x-3)}, \text { and } \\
\cos \angle A D C=\frac{D Q^{2}+D C^{2}-Q C^{2}}{2 \cdot D Q \cdot D C}=\frac{4 x^{2}+(4 x-3)^{2}-9}{2 \cdot 2 x(4 x-3)}
\end{gathered}
$$

From the equation

$$
\frac{9 x^{2}+(4 x-3)^{2}-13}{2 \cdot 3 x(4 x-3)}=\frac{4 x^{2}+(4 x-3)^{2}-9}{2 \cdot 2 x(4 x-3)}
$$

we find that $x=2$ or $x=\frac{2}{5}$, but for $x=\frac{2}{5}$, we get $4 x-3<0$. Therefore, $B C=x=2, Q D=2 x=4, C D=$ $4 x-3=5$

Since $Q C^{2}+Q D^{2}=9+16=25=D C^{2}$, triangle $C Q D$ is a right triangle, so $C Q=3$ is the height of the trapezoid. Therefore, the radius of the circle inscribed in the trapezoid is $\frac{3}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-42.jpg?height=429&width=740&top_left_y=1132&top_left_x=656)

## Answer

$\frac{3}{2}$.