[  The product of the lengths of the segments of chords and the lengths of the segments of secants [Theorem on the lengths of the tangent and secant; the product of the entire secant and its external part

 A circle inscribed in triangle $A B C$ divides the median $B M$ into three equal parts. Find the ratio $B C: C A: A B$.
#
Let $K, L$ and $N$ be the points of tangency of the inscribed circle with the sides $AC, BC$ and $AB$ respectively; $F$ and $Q$ be the points of intersection of the circle with the median $BM (F$ between $B$ and $Q)$. Suppose that point $K$ is located between points $M$ and $C$. Denote $BC=a, BF=FQ=QM=x$. Then

$BL^2 = BQ \cdot BF = 2 \times 2$.

Therefore, $BL = x\sqrt{2}, BN = BL = x\sqrt{2}$. Similarly, we find that $KM = x\sqrt{2}$, and since $CL = CK$, then

$$
MC = BC = a, AC = 2a, AB = AN + NB = a + x\sqrt{2} + x\sqrt{2} = a + 2x\sqrt{2}.
$$

Express the median $BM$ in terms of the sides of triangle $ABC$:

$$
4BM^2 = 2 \cdot BC^2 + 2 \cdot AB^2 - AC^2, \text{ or }
$$

$$
36 \times 2 = 2a^2 + 2(a + 2x\sqrt{2})^2 - 4a^2.
$$

From this equation, we find that $a = \frac{5x\sqrt{2}}{4}$. Then

$$
\begin{gathered}
BC = a = \frac{5x\sqrt{2}}{4}, AC = 2a = \frac{5x\sqrt{2}}{2} \\
AB = a + 2x\sqrt{2} = \frac{13x\sqrt{2}}{4}
\end{gathered}
$$

Therefore, $BC : CA : AB = 5 : 10 : 13$.

## Answer

$5 : 10 : 13$