XXIV OBM

Since the perimeter of $ADM$ and $ABCM$ are equal and share the side $AM$, $a+b+m=n+c⟺n-m=a+b-c$. 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 $$\begin{array}{rlrl}\frac{cn\sin (180^{\circ }-\theta )}{2}& =\frac{1}{2}\cdot \frac{ab+c(m+n)}{2}& & ⟺2cn=ab+cm+cn\\ ⟺c(n-m)& =ab& & ⟺c(a+b-c)=ab\\ ⟺(c-a)(c-b)& =0& & ⟺c=a\text{ or }c=b\end{array}$$ and we are done.