Let $P$ be the intersection of the tangent at $B$ to $\Gamma$ and the line $AD$. Then we have
$$\angle ABP = \angle ACB = \angle BAC.$$
This shows $BP // CA$. Note that $CB // AP$ since $ABCD$ is an isosceles trapezoid. Therefore, $PBCA$ is a parallelogram. As $N$ is the midpoint of $AB$, the diagonal $CP$ passes through $N$.

Next, the areas of $\triangle CDE$ and $\triangle CAE$ are the same since $MD = MA$. This implies
$$\frac{1}{2} DC \times DE \sin \angle CDE = \frac{1}{2} AC \times AE \sin \angle CAE.$$
As $\angle CDE + \angle CAE = 180^\circ$, we have $\sin \angle CDE = \sin \angle CAE$. Therefore,
$$\frac{ED}{EA} = \frac{CA}{CD} = \frac{BD}{BA}.$$
This shows $ABDE$ is a harmonic quadrilateral, and hence the tangent at $E$ to $\Gamma$ passes through $P$. This proves the tangent at $E$, the line $AD$ and the line $CN$ are concurrent at $P$.