Yes, a nice problem. In fact, the lines AM and AN cannot divide the angle BAD into three equal angles. Here is my proof: Assume that the lines AM and AN divide the angle BAD into three equal angles, i. e. assume that we have t = < BAM = < MAN = < NAD. Then, $\measuredangle BAD=\measuredangle BAM+\measuredangle MAN+\measuredangle NAD=3t$. Since < BAM = < NAD, and also < ABM = < ADN (this is because AB || DN and BM || AD), the triangles ABM and ADN are similar, so that $\displaystyle \frac{AB}{BM}=\frac{AD}{DN}$. Since BM = AD / 2 and DN = AB / 2, we can rewrite this as $\displaystyle \frac{AB}{AD}=\frac{AD}{AB}$, hence AB = AD, and our parallelogram ABCD is a rhombus. Therefore, $\measuredangle BAC=\measuredangle CAD=\frac{1}{2}\measuredangle BAD=\frac{1}{2}\cdot 3t=\frac{3}{2}t$. Hence, $\measuredangle MAC=\measuredangle BAC-\measuredangle BAM=\frac{3}{2}t-t=\frac{1}{2}t$. Also, since BM || AD, we have $\measuredangle ABM=180^{\circ }-\measuredangle BAD=180^{\circ }-3t$, and, since the triangle ABC is isosceles, we have $\measuredangle BCA=\measuredangle BAC=\frac{3}{2}t$, or, in other words, $\measuredangle MCA=\frac{3}{2}t$. Thus, by the Sine Law in triangle ABM, we find $\displaystyle \frac{BM}{AM}=\frac{\sin \measuredangle BAM}{\sin \measuredangle ABM}=\frac{\sin t}{\sin \left( 180^{\circ }-3t\right) }=\frac{\sin t}{\sin \left( 3t\right) }$, and by the Sine Law in triangle ACM, we find $\displaystyle \frac{CM}{AM}=\frac{\sin \measuredangle MAC}{\sin \measuredangle MCA}=\frac{\sin \left( \frac{1}{2}t\right) }{\sin \left( \frac{3}{2}t\right) }$. Now, since M is the midpoint of BC, we have BM = CM, so that the left hand sides of the last two equations are equal. Therefore we conclude $\displaystyle \frac{\sin t}{\sin \left( 3t\right) }=\frac{\sin \left( \frac{1}{2}t\right) }{\sin \left( \frac{3}{2}t\right) }$, or, equivalently, $\displaystyle \frac{\sin t}{\sin \left( \frac{1}{2}t\right) }=\frac{\sin \left( 3t\right) }{\sin \left( \frac{3}{2}t\right) }$. Using the formula $\sin x=2\sin \left( \frac{1}{2}x\right) \cos \left( \frac{1}{2}x\right)$, this becomes $\displaystyle \cos \left( \frac{1}{2}t\right) =\cos \left( \frac{3}{2}t\right) $. But $\frac{1}{2}t$ and $\frac{3}{2}t$ are acute angles (since 3t = < BAD is an angle in our parallelogram!). Hence, we have $\frac{1}{2}t=\frac{3}{2}t$, what is impossible. Proof complete. Darij

Actually, I also found this solutiuon (alike) but I'm still a beginner (13 years) so I posted it to see if my solution is correct, or to see another solution. Thanks.