The right angle $DCA$ subtends the chord $AD$ of the circumcircle of triangle $ACD$ (Fig. 26), therefore $AD$ is a diameter of this circle. Since point $K$ lies on the same circle, $\angle AKD = 90^\circ$. From supplementary angles we get $\angle MKA = 180^\circ - \angle AKD = 90^\circ$.

From the conditions of the problem $AM = MB$ and $KM = ML$ and from the vertex angles we get $\angle AMK = \angle BML$. Thus, the triangles $AMK$ and $BML$ are equal. Consequently $\angle MLB = \angle MKA = 90^\circ$.

Next, we show that $\angle NLK = \angle NKA$. It suffices to show the equality $\angle NLB = \angle NKD$ since $\angle NLK = \angle NLB - \angle KLB = \angle NLB - 90^\circ$ and

$\angle NKA = \angle NKD - \angle AKD = \angle NKD - 90^\circ$. Furthermore, since $N, L, B$ and $C$ lie on the same circle in this order, we have $\angle NLB = 180^\circ - \angle BCN$, while the points $N, K, D$ and $C$ lying on one circle in this order implies $\angle NKD = 180^\circ - \angle DCN = 180^\circ - \angle BCN$. So $\angle NLK = \angle NKA$ as desired. Hence $\angle NLK + \angle LKN = \angle NLK + \angle LKA - \angle NKA = \angle LKA = 90^\circ$, which implies $\angle KNL = 180^\circ - 90^\circ = 90^\circ$.

As in Solution 1 we show that $AD$ is a diameter of the circumcircle of triangle $ACD$ and $\angle MLB = 90^\circ$. Since point $N$ lies on the circumcircle of triangle $ACD$ we have $\angle AND = 90^\circ$.

Let $T$ be the intersection of the lines $LD$ and $AC$ (Fig. 27). From supplementary angles we get $\angle TCB = 180^\circ - \angle BCA = 90^\circ$. Since $\angle TLB = 90^\circ$, points $C$ and $L$ lie on the circle with the diameter $TB$, whence $T$ lies on the circumcircle of triangle $LBC$.

We show that $\angle DNL = \angle ANK$. Since the points $C, N, L, T$ lie on the same circle in this order, we have $\angle CNL + \angle LTC = 180^\circ$. Hence
$$\angle DNL = \angle CNL - \angle CND = 180^\circ - \angle LTC - \angle CND \\ = 180^\circ - \angle DTA - \angle TAD.$$
From triangle $TAD$ we get $180^\circ - \angle DTA - \angle TAD = \angle ADT$ and from supplementary angles $\angle ADT = 180^\circ - \angle KDA$. Since $A, N, K, D$ lie on the same circle in this order, we get $180^\circ - \angle KDA = \angle ANK$. So indeed $\angle DNL = \angle ANK$.

Finally notice that $\angle ANK = \angle AND + \angle DNK = 90^\circ + \angle DNK$, hence $\angle DNL = 90^\circ + \angle DNK$. Consequently $\angle KNL = \angle DNL - \angle DNK = 90^\circ$.