Estonian Mathematical Olympiad

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 $\angle KNL=180^{\circ }-90^{\circ }=90^{\circ }$.

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Alternative solution.

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 $$\begin{gathered} \angle DNL=\angle CNL-\angle CND=180^{\circ }-\angle LTC-\angle CND \\ =180^{\circ }-\angle DTA-\angle TAD. \end{gathered}$$ 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 $\angle KNL=\angle DNL-\angle DNK=90^{\circ }$.