First we prove that $N L \perp M C$. The arguments depend slightly on the position of $D$. The other cases are similar.

From the cyclic quadrilaterals $A D C M$ and $D N B C$ we have:

$$\angle D C L=\angle D A M \text { and } \angle C D L=\angle C B N \text {. }$$

So we obtain

$$\angle D C L+\angle C D L=\angle D A M+\angle C B N .$$

And because $A D \| B C$, if $Z$ the point of intersection of $A M, B C$ then $\angle D A M=\angle B Z A$, and we have

$$\angle D C L+\angle C D L=\angle B Z A+\angle C B N=90^{\circ}$$

Let $P$ the point of intersection of $K L, A C$, then $N P \perp A C$, because the line $K P L$ is a Simson line of the point $N$ with respect to the triangle $A C M$.

From the cyclic quadrilaterals $N P C L$ and $A N D C$ we obtain:

$$\angle C P L=\angle C N L \text { and } \angle C N L=\angle C A D \text {, }$$

so $\angle C P L=\angle C A D$, that is $K L\|A D\| B C$ therefore $\angle K T C=\angle A D C$ (1).

But $\angle A D C=\angle A N C=\angle A N K+\angle K N C=\angle C N L+\angle K N C$, so

$$\angle A D C=\angle K N L$$

From (1) and (2) we obtain the result.

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