, 2008

Solution:

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:
$$\text{D C L = D A M and C D L = C B N.}$$
So we obtain
$$\text{D C L + C D L = D A M + C B N.}$$
And because $A D \parallel B C$, if $Z$ is the point of intersection of $A M$, $B C$ then $\text{D A M = B Z A}$, and we have
$$\text{D C L + C D L = B Z A + C B N = 90}$$
Let $P$ be 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:
$$\text{C P L = C N L and C N L = C A D,}$$
so $\text{C P L = C A D}$, that is $K L \parallel A D \parallel B C$ therefore $\text{K T C = A D C}$ (1).

But $\text{A D C = A N C = A N K + K N C = C N L + K N C}$, so
$$\text{A D C = K N L}$$
From (1) and (2) we obtain the result.