3. Points $K, L$ and $M$ are the midpoints of sides $AB, BC$ and $CD$ of parallelogram $ABCD$. It turned out that quadrilaterals $KBLM$ and $BCDK$ are cyclic. Find the ratio $AC: AD$.

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The translation is provided as requested, maintaining the original formatting and structure.
Answer: 2.

![](https://cdn.mathpix.com/cropped/2024_05_06_66368caef28b3de8e403g-15.jpg?height=417&width=452&top_left_y=1887&top_left_x=882)

Solution. Let $N$ be the point of intersection of the diagonals of the parallelogram. Quadrilateral $K B L M$ is an inscribed trapezoid, so $K B = L M$. Multiplying this equality by 2, we get $A B = B D$. Then $A K = D N$ and

$$
\angle A K N = 180^{\circ} - \angle K A D = 180^{\circ} - \angle A D N = \angle D N K
$$

Thus, triangles $A K N$ and $D N K$ are equal, from which $D K = A N$. On the other hand, $B C D K$ is also an inscribed trapezoid, so $D K = B C = A D$. Then

$$
\frac{A C}{A D} = \frac{2 A N}{A D} = \frac{2 A N}{D K} = 2
$$