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!
 A: Hint 1: triangles ABM and CBM must be similar. (And the equivalent triangles on the other side).
Hint 2: 

 Angle AEB = angle CED 

Hint 3:

 So angle ECN = angle EBM

A: You may refer to the following diagram:

Consider $ΔABE$ and $ΔEDC$.
$ΔABE \sim ΔEDC$ as they have equal corresponding angles.
As you propose, construct a line from $D$ meeting $EC$ at $P$, parallel to $BM$.
$DP$, as $BM$ does, bisects $∠EDC$ and cuts $EC$ in half.  
Then $NC$ and $DP$ bisect $∠DCE$ and $∠EDC$, and halve $ED$ and $EC$, respectively.
Hence $ΔEDC$ is $60$-$60$-$60$.
Why? Construct a line from $E$ meeting $DC$ at $Q$ through the intersection of $NC$ and $DP$. By sine formula, 
$$\frac{\sin∠EQC}{EC}=\frac{\sin∠CEQ}{QC} \text{ and } \frac{\sin∠DQE}{ED}=\frac{\sin∠QED}{DQ}$$
$$\frac{\sin∠CEQ}{QC}=\frac{\sin∠QED}{DQ}\Longrightarrow \frac{\sin∠EQC}{EC}=\frac{\sin∠DQE}{ED}\Longrightarrow EC=ED.$$
Using the same argument, we can infer that $ΔEDC$ has equal sides.
Could you prove that $∠ADC=90^\circ$?
One interesting thing is that the intersecting point of $NC$ and $DP$ is both the incenter and the centroid of $ΔEDC$. Dr. Floor, from Ask Dr. Math, proved that a triangle being equilateral, and the centroid being incenter, is biconditional   (Equilateral Triangle - Centroid/Incenter).
A: 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. 
A: $\underline{\text{ Another Method:}}$
Please refer to the following figure:

Consider $ΔAED$.
By the midpoint theorem, $MN \parallel AD$ and $MN=\frac{1}{2}AD$.
Now extend $BM$ and $CN$ to meet one another at $F$, and consider $ΔBFC$, $ΔABE$ and $ΔEDC$.
$$∠FBC=∠ABE=∠EDC \text{ and } ∠FCB=∠EAB=∠ECD $$
$$\Longrightarrow ΔBFC \sim ΔABE \sim ΔEDC.$$
As $CN$ bisects $\angle ECD$ and halves $ED$, $MC$ bisects $\angle FCB$ and halves $FB$; as $BM$ bisects $\angle ABE$ and halves $AE$, $NB$ bisects $\angle FBC$ and halves $FC$.
That means $M$ and $N$ are the midpoints of $FB$ and $FC$, respectively. Hence, by the midpoint theorem, $MN\parallel BC$ and $MN=\frac{1}{2}BC$.
Thus, $AD\parallel BC$ and $AD=BC$. 
Then quadrilateral $ABCD$ is a parallelogram.
Consider $ΔABM$ and $ΔCBN$; $ΔCDN$ and $ΔCBM$.
$$∠ABM=∠CBN \text{ and } ∠NCD=∠MCB$$
$$∠MAB=∠NCB \text{ and } ∠CDN=∠CBM$$
$$\Longrightarrow ΔABM \sim ΔCBN \text{ and } ΔCDN \sim ΔCBM.$$
$$\frac{AB}{DC}=1 \implies \frac{AB}{BC}=\frac{DC}{BC}. $$
That means the ratio of the corresponding sides of $ΔABM$  and $ΔCBN$ is the ratio of the corresponding sides of $ΔCDN$ and $ΔCBM $. Hence
$$\frac{AM}{NC}=\frac{NC}{MC} \implies 3AM^2=NC^2$$
$$\implies NC=\sqrt{3}AM $$
Consider $ΔEDC$.
Using the angle bisector theorem, as Rex proposed,
$$\frac{CE}{CD}=\frac{EN}{ND}=1 \implies CE=CD \implies CN \perp ED.$$
Consider $ΔENC$.
Since $EC=2AM$ and $NC=\sqrt{3}AM$ and $CN \perp ED$, $EN=\sqrt{4-3}AM=AM.$
Then $ΔENC$ is $30$-$60$-$90$. Thus, $ΔAEB$ and $ΔEDC$ are $60$-$60$-$60$. Since $AD\parallel BC$, quadrilateral $ABCD$ is a rectangle.
