Saudi Arabian Mathematical Competitions

Let $A'$, $B'$, $C'$ be the midpoints of sides $BC$, $CA$, $AB$, respectively. Construct the parallelogram $AC'CC''$. The points $C'$, $B'$, $C''$ are collinear, hence $BA'C''B'$ is also a parallelogram, that is $A'C'' = BB'$.

The desired triangle is $AA'C''$, and $AA' = m_{a}$, $A'C'' = m_{b}$, $AC'' = m_{c}$.

b. Recall the median formula
$$m_{a}^{2} = \frac{1}{4}\left(2(b^{2} + c^{2}) - a^{2}\right)$$
together with the other two similar formulas for $m_{b}$ and $m_{c}$.
Assume that the squares of the sides are in arithmetic progression, for example $2b^{2} = a^{2} + c^{2}$. Then $m_{a}^{2} = \frac{3}{4}c^{2}$, $m_{b}^{2} = \frac{3}{4}b^{2}$, $m_{c}^{2} = \frac{3}{4}a^{2}$, implying that
$$\frac{m_{a}}{c} = \frac{m_{b}}{b} = \frac{m_{c}}{a} = \frac{\sqrt{3}}{2}$$
Obviously, the triangles are similar.
Conversely, assume the triangles are similar. Let $a \leq b \leq c$. Then $m_{a} \geq m_{b} \geq m_{c}$, so
$$\frac{m_{a}}{c} = \frac{m_{b}}{b} = \frac{m_{c}}{a} = k$$
From the equality $m_{a}^{2} + m_{b}^{2} + m_{c}^{2} = \frac{3}{4}(a^{2} + b^{2} + c^{2})$ we get $k = \frac{\sqrt{3}}{2}$. Hence $4m_{b}^{2} = 3b^{2}$, which yields $2b^{2} = a^{2} + c^{2}$.