Denote by $a$, $b$, $c$ the sidelengths, and by $m_a$, $m_b$, $m_c$ the lengths of the medians of the triangle $ABC$. Since $MD$ is median in the right-angled triangle $AMC$, it follows that $2m_b/3 = MD = AD = CD = b/2$, so $m_b = 3b/4$, whence $(3b/4)^2 = m_b^2 = (a^2 + c^2)/2 - b^2/4$; that is, $13b^2 = 8(a^2 + c^2)$.

Next, apply the Menelaus theorem to get $EC/EB = 4 = FA/FB$ and deduce thereby that the lines $AC$ and $EF$ are parallel. The quadrangle $AFED$ is therefore a trapezium; it is cyclic if and only if $AF = DE$.

Express the two in terms of $a$, $b$ and $c$. Recall that $FA/FB = 4$ to obtain $AF = 4c/5$. Next, apply Stewart's theorem in triangle $BCD$ to get $DE^2 = b^2/2 - 4a^2/25$. By the preceding, the quadrangle $AFED$ is cyclic if and only if $25b^2 - 8a^2 = 32c^2$. Recall that $13b^2 = 8(a^2 + c^2)$ to express $b$ and $c$ in terms of $a$: $b = 2a\sqrt{2}/3$ and $c = 2a/3$.

Finally, let $N$ be the midpoint of the side $BC$ and let the lines $AN$ and $EF$ meet at $P$. Notice that $EN = a/2 - a/5 = 3a/10$, and the triangles $ANC$ and $PNE$ are similar, to obtain $NP = 3m_a/5$, so
$$NA \cdot NP = 3m_a^2/5 = 3(2(b^2 + c^2) - a^2)/20 = a^2/4 = NB \cdot NC.$$
The conclusion follows.