First, observe that $\angle AED = \angle FCA < \angle BCA = \angle EAD$ so $AD < DE$. So if it is not possible to make a triangle from the segments, then we must have $DE > DA + AF$. Note that $\triangle AFC \sim \triangle ADE$. This gives $\frac{AF}{AD} = \frac{FC}{DE} = \frac{AC}{AE}$ so $AF = \frac{AC \cdot AD}{AE}$ and $DE = \frac{AE \cdot CF}{AC}$, so the above statement becomes: $$\frac{AE \cdot CF}{AC} > \frac{AD(AC+AE)}{AE}$$which rearranges to $AE^2 \cdot CF > (AC+AE) \cdot AD \cdot AC = (AC+AE) \cdot AE \cdot AF$ so we get $AE \cdot CF > (AC+AE) \cdot AF$. Let $K$ be the point on $AC$ such that $FA = FC$. Since $F$ is beyond $M$, it follows that $K$ is closer to $C$ than to $A$. Then the equation becomes $AC \cdot AF < AE(CF-AF) = AE(FC-FK) < AE \cdot CK < \frac{AE \cdot AC}{2}$ giving $BC < 2 \cdot AF < AE < BC$, a contradiction. Therefore, the three given sides do form a triangle, as desired. $\blacksquare$