Brazilian Mathematical Olympiad

Since the perimeter of $ADM$ and $ABCM$ are equal and share the side $AM$, $a + b + m = n + c \iff n - m = a + b - c$. Since they also have the same area, the area of $ADM$ is half the area of $ABCD$. If $\angle ABC = \theta$, then $\angle CDA = 180^\circ - \theta$ and
$$\begin{aligned} \frac{c n \sin(180^\circ - \theta)}{2} &= \frac{1}{2} \cdot \frac{a b + c(m + n)}{2} &&\iff 2 c n = a b + c m + c n \\ \iff c(n - m) &= a b &&\iff c(a + b - c) = a b \\ \iff (c - a)(c - b) &= 0 &&\iff c = a \text{ or } c = b \end{aligned}$$
and we are done.