Solution:

Let $I$ be the incentre of triangle $ABC$ and $D$ be its projection on $BC$. Observe that $AB \neq AC$ as $AB = AC$ implies that $D = L = M$. So assume that $AC > AB$. Let $N$ be the projection of $I$ on $KL$. Then the perpendicular $IN$ from $I$ to $KL$ is a bisector of $KL$ and as $AK = LM$, it is a bisector of $AM$ also. Hence $AI = IM$.

Fig. 1.

But $AI = \frac{r}{\sin(A/2)} = r \operatorname{cosec}(A/2)$ and
$$\begin{aligned} IM^2 & = ID^2 + DM^2 = r^2 + (BM - BD)^2 \\ & = r^2 + \left(\frac{a}{2} - (s-b)\right)^2 \end{aligned}$$
Hence $r^2 \operatorname{cosec}^2(A/2) = r^2 + \left((a/2) - (s-b)\right)^2$ giving $r^2 \cot^2(A/2) = ((b-c)/2)^2$. Since $b > c$, we obtain $r \cot(A/2) = ((b-c)/2)$. So $s-a = ((b-c)/2)$. This gives $a = 2c$.

As $KN = NL$ and $AK = KL = LM$, we have $NL = AM/6$. We also have $AN = NM$. Now
$$\begin{aligned} r^2 = IL^2 = IN^2 + NL^2 & = AI^2 - AN^2 + NL^2 \\ & = AI^2 - \frac{1}{4} m_a^2 + \frac{1}{36} m_a^2 \\ & = r^2 \operatorname{cosec}^2(A/2) - \frac{2}{9} m_a^2 \end{aligned}$$
Hence $r^2 \cot^2(A/2) = \frac{2}{9} m_a^2$. From the above, we get
$$\left(\frac{b-c}{2}\right)^2 = \frac{2}{9} \cdot \frac{1}{4} (2b^2 + 2c^2 - a^2)$$
Simplification gives $5b^2 + 13c^2 - 18bc = 0$. This can be written as $(b-c)(5b-13c) = 0$. As $b \neq c$, we get $5b - 13c = 0$. To conclude, $a = 2c$, $5b = 13c$ yield
$$\frac{a}{10} = \frac{b}{13} = \frac{c}{5}$$