Iranian Mathematical Olympiad

First, we prove that $KB=LC$.

Let $F$ and $E$ be the feet of perpendiculars from $K$ and $L$ to $MB$ and $MC$, respectively. We have $$OK=MB\Rightarrow KF=\text{The distance from }O\text{ to }MB.$$ $$OL=MC\Rightarrow LE=\text{The distance from }O\text{ to }MC.$$ Since $MB=MC$, $O$ (circumcenter of triangle $MBC$) is equidistant from these equal chords, and therefore $KF=LE$ (1). On the other hand, since $ABMC$ is cyclic, $\angle KBF=\angle LCE$ (2). (1) and (2) imply that the two right-angled triangles $KBF$ and $LCE$ are congruent, so their hypotenuses have the same length ($KB=LC$). Let $D$ be the midpoint of arc $B\bar{A}C$. We have $$\begin{array}{l}DB=DC\\ \angle DBA=\angle DCA\\ KB=LC\end{array}\}\Rightarrow \stackrel{△}{\mathrm{DBK}}\equiv \stackrel{△}{\mathrm{DCL}}\Rightarrow \{\begin{array}{l}\angle KDL=\angle BDC=\angle BAC\\ KD=LD\end{array}$$ Since $\angle KOL=\angle BMC$, we deduce that $AKOLD$ is cyclic. Therefore, $$\begin{gathered} \begin{array}{l}\angle AKD=\angle ALD=\angle AOD=\angle B-\angle C\\ \angle KBD=\angle LCD=\frac{\angle B-\angle C}{2}\end{array}\}\Rightarrow \angle KDB=\angle LDC=\frac{\angle B-\angle C}{2} \\ \Rightarrow KB=KD=LD=LC\text{ (3).} \end{gathered}$$



$$\begin{array}{rl}\angle NDK& =\angle NDB+\angle BDK=(90^{\circ }-\angle B)+\left(\frac{\angle B-\angle C}{2}\right)=\frac{\angle A}{2}\\ \angle NDL& =\angle NDC-\angle CDL=(90^{\circ }-\angle C)-\left(\frac{\angle B-\angle C}{2}\right)=\frac{\angle A}{2}\end{array}\}$$ $$\Rightarrow \angle NDK=\angle NDL(4).$$ (3) and (4) imply that two triangles NDK and NDL are congruent. Therefore, we have $NK=NL$.