3. In triangle $A B C \quad A C=18 \text{ cm}, B C=21$ cm. Point $K$ is the midpoint of side $B C$, and point $M$ is the midpoint of side $A B$, point $N$ lies on side $A C$ and $A N=6 \text{ cm}$. Given that $M N=K N$. Find the length of side $A B$.
3. Since $M$ and $K$ are midpoints of the sides, we will extend the segments $NM$ and $NK$ beyond the specified points by the same distance and connect the points $L, B, A, N$; as well as $F, B, N, C$. Then the quadrilaterals $ALBN$ and $NBFC$ become parallelograms. Since in a parallelogram the sum of the squares of the diagonals is equal to the sum of the squares of its sides, we have:

$NF^2 + BC^2 = 2NB^2 + 2NC^2$ (for parallelogram $NBFC$) and $AB^2 + NL^2 = 2AN^2 + 2NB^2$ (for parallelogram $ALBN$).

From the first equation, express $NF^2 = 2NB^2 + 288 - 441$ and, considering that $MN = KN$, we get: $AB^2 + 2NB^2 - 153 = 72 + 2NB^2$, from which $AB = 15 \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2d68f1e45d2f307462c1g-1.jpg?height=517&width=1251&top_left_y=2083&top_left_x=277)

Answer: 15 cm