Given a specific trapezoid, prove it is a rectangle > In quadrilateral ABCD, AB is parallel to CD. AC and BD meet at E. Points M and N are the midpoints of AE and DE, respectively. BM and BE trisect $\angle ABC$, and CE and CN trisect$ \angle BCD$. Prove that ABCD is a rectangle. I noted that $\angle ABE $ and $\angle EDC$ are congruent and constructed DP, the angle bisector of $\angle EDC$. I also thought about using angle addition to show the angles. I think angle addition and the angle sum of a triangle is the way to go, but I don't know how to implement it. It might be something along the lines of 30-60-90 triangles. How can I continue? Please provide some hints to the end (as I struggled with this question for over two hours already). Thanks!

First, we can determine that the trisected angles are all congruent. Because of this, $\triangle$${ECD}$'s median CN is an angle bisector of $\triangle$${ECD}$. The angle bisector theorem tells us that $\frac{EC}{CD} = \frac{EN}{ND} = 1$, therefore $\triangle{ECD}$ is isosceles. Because of this, we can find that AB = BE and BM is perpendicular to AE. $\angle{BME} = \angle{CNE}$ and $\angle{MEB} = \angle{CEN}$, $\triangle{BEM} \sim \triangle{CEN}$. Therefore, $\angle{ECN} = \angle{EBM}$, and $\angle{ABC} = 3\angle{MBE} = 3\angle{ECN} = \angle{BCD}$. $\angle{ABC} + \angle{BCD} = 180$ degrees. Therefore, each of those two angles are right angles. Since $\angle{EBC}=\angle{ECB}$, $BE = CE$. The isosceles triangles show that $CD = CE = BE = BA$. ABCD is a parallelogram since there's a pair of parallel and congruent opposite sides. Since $\angle{ABC}$ is right, $ABCD$ is a rectangle.