$ALOK$ is cyclic($\angle KOL=\angle CMB=180-\angle KAL$) with now let $AN$ meet circle $AKL$ at $S$. Since $\angle LAS=90-\angle CBA=\angle OAK$ then $OK=LS$ and $SOKL$ is an isosceles trapezoid with the bisector of $SO$ and $KL$ coinciding. $\angle OSA=\angle OLA=\angle MBA=\angle MNA$ so $MN||OS$ but $OM||SN$ so $OMNS$ is a parallelogram and $NS=OM=ON$ so $N$ is on the bisector of $OS$ so $NK=NL$.

i've got a metric proof for this : using cosines lemma, use that CL=BK=(a)/(2(cos B+ cos C))

cosines lemma on triangles : CLN and BKN

another trick is that AKLOT is cyclic where T is the anti-diameter of M.

We use complex numbers. Denote by the lowercase letter of a point its complex coordinate. Let the circumcircle of $ABC$ be the unit circle, and let $OM$ be the real axis. Clearly $m = 1$ and $b = \bar{c} = \frac{1}{c}$. It is well-known that $N$ is the reflection of $H$ over $BC$, where $H$ is the orthocenter of $ABC$. Therefore, we have $\frac{h+n}{2} = \text{proj}_{BC} h$, or $h+n = b+c+h-bc\bar{h}$, so $n = b+c-bc\bar{h} = b + c - \frac{bc}{a} - c - b = \frac{bc}{a}$. Now $a, l, c$ are collinear and $OL \parallel MC$, so \[\frac{l-c}{\bar{l}-\bar{c}} = -ac\] and \[\frac{l}{\bar{l}} = \frac{c-m}{\bar{c}-\bar{m}} = \frac{c-1}{\frac{1}{c} - 1} = -c\] Therefore $l = -c\bar{l}$, so we have $\frac{l-c}{-\frac{l}{c} - \frac{1}{c}} = -ac \implies l = \frac{a+c}{1-a}$. Similarly, $k = \frac{a+b}{1-a}$. Now we have to prove that $NK^2 = NL^2$. Since \begin{align*}NK^2 &= (n-k)(\bar{n}-\bar{k}) \\ & = \left(\frac{bc}{a} - \frac{a+b}{1-a}\right)\left(\frac{\frac{1}{bc}}{\frac{1}{a}} - \frac{\frac{1}{a}+\frac{1}{b}}{1-\frac{1}{a}}\right) \\ &= \left(\frac{bc - abc - a^2 -ab}{a(1-a)}\right)\left(\frac{a^2b - ab - abc - b^2c}{b^2c(a-1)}\right)\end{align*} and \begin{align*}NL^2 &= (n-l)(\bar{n}-\bar{l}) \\ &= \left(\frac{bc}{a} - \frac{a+c}{1-a}\right)\left(\frac{\frac{1}{bc}}{\frac{1}{a}} - \frac{\frac{1}{a}+\frac{1}{c}}{1 - \frac{1}{a}}\right) \\&= \left(\frac{bc-abc-a^2-ac}{a(1-a)}\right)\left(\frac{a^2c - ac - abc - bc^2}{bc^2(a-1)}\right)\end{align*} we have to prove that $c(bc-abc-a^2-ab)(a^2b-ab-abc-b^2c) = b(bc-abc-a^2-ac)(a^2c-ac-abc-bc^2)$, and using the fact that $bc = 1$, we see that this is true.

it s easy to now that we have point S on BC with KSB=ABC and LSC=ACB because AKOL is cyclic.now we most prove that NSK=NSL. now with use BK=CL prove BNS=CNS (BN/CN=BS/CS).NSK=180-A/2,NSL=180-A/2.

Moreover, if we take $F$ to be the midpoint of big arc BC, we have that $NF$ is the perpendicular bisector of KL.

why $B$??? Small mistake.

My solution is the same as $MBGO$. But $KC=LB=\frac{R\cos{\frac{a}{2}}a}{\sin{\frac{a}{2}}(c+b)}$

My solution is the same as $MBGO$. But $KC=LB=\frac{R\cos{\frac{a}{2}}a}{\sin{\frac{a}{2}}(c+b)}$

dan23 wrote: Let $M$ be the midpoint of (the smaller) arc $BC$ in circumcircle of triangle $ABC$. Suppose that the altitude drawn from $B$ intersects the circle at $N$. Draw two lines through circumcenter $O$ of $ABC$ paralell to $MB$ and $MC$, which intersect $AB$ and $AC$ at $K$ and $L$, respectively. Prove that $NK=NL$. Moderator please correct the mistake , the altitude is drawn from vertex $A$ , not $B$. Done. Only eight months late... -- v_E 4/22/2014

leader wrote: let $AN$ meet circle $AKL$ at $S$. whT DOES IT MEAN?

It means that $S$ is the intersection point of line $AN$ and circumcircle of $AKL$

Obviously $AM$ is the angle bisector of $\angle ABC$. As $\angle ALO + \angle AKO = \angle ACM + \angle ABM = \angle C + \angle \frac A2 + \angle \frac A2 + \angle B = \pi$ we have that $A,K,O,L$ are cyclic. Now let $P$ be the second intersection of $AM$ and $\odot AKOL$. Also let $AM$ intersect $BC$ at $Q$. Now we have: $$\angle APO = \angle AKO = \angle ABM = \angle B + \angle \frac A2 = \pi - \angle AQB = \angle AQC$$ Becasue of this it follows that $OP \parallel BC \implies OP \perp AN$. And as $OA=ON$ it follows that $OP$ is the perpendicular bisector of $AN$. Now using this and that $\triangle APO \sim \triangle NPO$ it follows that: $\angle PNO = \angle PAO = \angle PMO$. This implies that $N,M,O,P$ are cyclic. Let $X$ be the antipodal point of $M$, and $D$ the antipodal of $A$ in $\odot ABC$. Now using isogonality we have: $\angle PNM = \angle PND + \angle DNM = \angle PND + \angle DAM = \angle PND + \angle NAM = \angle PND + \angle PNA = \angle AND = \frac{\pi}2$. On the other side obviously $\angle XNM = \frac{\pi}2$, therefore $X,P,N$ are colinear. Now we have: $\angle PXO = \angle PMO = \angle PAO \implies X \in \odot ALOK$. Finally as $\angle XAP = \angle XAD = \frac{\pi}2$, it follows that $XP$ is a diameter in $\odot ALOK$ and hence it bisects $LK$. As $N$ lies on the perpendicular bisector of $LK$, we have that $NK=NL$. Q.E.D.

It is easy to see that $AKOL$ is cyclic. Let $OM$ cut the circumcircle of $ABC$ at $F$. Then $\angle AFM = \angle ACM = \angle ALO$, so $F$ lies in the same circle, and $\angle LKF = \angle CAF = \angle CMF = \angle FMB = 180 - \angle KAF = \angle FLK$, so $F$ belongs to the perpendicular bisector of $LK$. We will see that $NF$ is perpendicular to $LK$, so it must be its bisector. Since $NF$ is perpendicular to $NM$, we have to show that this is parallel to $LK$. Using that $OK$ and $BM$ are parallel and $\angle OKL = \angle OAC = 90 - \angle ABC = \angle BAN = \angle BMN$, the result follows.

dan23 wrote: Let $M$ be the midpoint of (the smaller) arc $BC$ in circumcircle of triangle $ABC$. Suppose that the altitude drawn from $A$ intersects the circle at $N$. Draw two lines through circumcenter $O$ of $ABC$ paralell to $MB$ and $MC$, which intersect $AB$ and $AC$ at $K$ and $L$, respectively. Prove that $NK=NL$. We will use complex numbers with unit circle $(ABC)$, $A=a$, $B=b$, $C=c$, $M=-\sqrt{bc}$. Step 1: We will compute $N$. Denote $N=n$ we know that: $$a \cdot n + b \cdot c = 0 \implies n=- \frac{bc}{a} \implies \overline{n}=-\frac{a}{bc}$$Step 2: We will compute $K$ and $L$. Denote $K=k$ and $L=l$ we will use the propety of parallel lines and colinear points on complex numbers, first for $K$ and later for $L$: $$\frac{a-k}{a-b}=\frac{\overline{a}-\overline{k}}{\overline{a}-\overline{b}}=\frac{\frac{1-\overline{k} \cdot a}{a}}{\frac{b-a}{ab}}=\frac{(\overline{k} \cdot a-1)b}{a-b} \implies a+b=\overline{k} \cdot a \cdot b+k$$Now we will use the parallel propety: $$\frac{k}{b+\sqrt{bc}}=\frac{\overline{k}}{\overline{b}+\overline{\sqrt{bc}}}=\frac{\overline{k} \cdot b \cdot \sqrt{bc}}{b+\sqrt{bc}} \implies \overline{k}=\frac{k}{b \cdot \sqrt{bc}}$$Both equations gives to you: $$a+b=k \left( \frac{a}{\sqrt{bc}}+1 \right) \implies k=\frac{(a+b)\sqrt{bc}}{a+\sqrt{bc}}$$Now lets compute $L$ : $$\frac{a-l}{a-c}=\frac{\overline{a}-\overline{l}}{\overline{a}-\overline{c}}=\frac{(\overline{l} \cdot a-1)c}{a-c} \implies a-l=\overline{l} \cdot a \cdot c -c \implies a+c=\overline{l} \cdot a \cdot c +l$$Now lets use the parallel propety: $$\frac{l}{c+\sqrt{bc}}=\frac{\overline{l}}{\overline{c}+\overline{\sqrt{bc}}}=\frac{\overline{l} \cdot c \sqrt{bc}}{c+\sqrt{bc}} \implies \overline{l}=\frac{l}{c \sqrt{bc}}$$Combine both equations to get: $$a+c=l \left(\frac{a}{\sqrt{bc}}+1 \right) \implies l=\frac{(a+c)\sqrt{bc}}{a+\sqrt{bc}}$$Step 3: Compute the conjugates of $k$ and $l$. Remeber when we used a propety that we get from the conjugates so lets use it!. $$\overline{k}=\left(\frac{(a+b)\sqrt{bc}}{a+\sqrt{bc}} \right) \frac{1}{b \sqrt{bc}}=\frac{a+b}{(a+\sqrt{bc})b}$$$$\overline{l}=\left(\frac{(a+c)\sqrt{bc}}{a+\sqrt{bc}} \right) \frac{1}{c \sqrt{bc}}=\frac{a+c}{(a+\sqrt{bc})c}$$Step 4: Make the midpoint of $KL$ ($U=u$) and make the feet of perpendicular from $N$ to $KL$ ($U'=u'$) and then prove that they are the same point: When we define the midpoint of $KL$ we get this: $$u=\frac{1}{2} \left(\frac{(a+b+a+c)\sqrt{bc}}{a+\sqrt{bc}} \right)=\frac{(2a+b+c)\sqrt{bc}}{2(a+\sqrt{bc})}$$When we define the feet of the perpendicular from $N$ to $KL$ we get: $$u'=\frac{(\overline{k}-\overline{l})n+(k-l)\overline{n}+\overline{k}l-k \overline{l}}{2(\overline{k}-\overline{l})}$$I will simplify the expresions part for part: $$\overline{k}-\overline{l}=\frac{a(c-b)}{bc(a+\sqrt{bc})} \implies (\overline{k}-\overline{l})n=\frac{b-c}{a+\sqrt{bc}}$$$$k-l=\frac{(b-c)\sqrt{bc}}{a+\sqrt{bc}} \implies (k-l)\overline{n}=\frac{a(c-b)}{\sqrt{bc}(a+\sqrt{bc})}$$$$\overline{k}l-k\overline{l}=\left(\frac{(a+b)(a+c)\sqrt{bc}}{(a+\sqrt{bc})^2} \right) \frac{c-b}{bc}=\frac{(a+b)(a+c)(c-b)}{(a+\sqrt{bc})^2\sqrt{bc}}$$Then the new expresion of $u'$ before simplifying $\frac{c-b}{a+\sqrt{bc}}$ is: $$u'=\frac{-1+\frac{a}{\sqrt{bc}}+\frac{(a+b)(a+c)}{(a+\sqrt{bc})\sqrt{bc}}}{\frac{2a}{bc}}=\frac{-bc(a+\sqrt{bc})+a\sqrt{bc} \cdot (a+\sqrt{bc})+(a+b)(a+c)\sqrt{bc}}{2a(a+\sqrt{bc})}$$This is actualy good we are getting closer and closer to the claimed result. $$u'=\frac{-abc-bc\sqrt{bc}+a^2\sqrt{bc}+abc+a^2\sqrt{bc}+a\sqrt{bc}(b+c)+bc\sqrt{bc}}{2a(a+\sqrt{bc})}=\frac{(2a+b+c)\sqrt{bc}}{2(a+\sqrt{bc})}=u$$This means that the bisector line of $KL$ passes throught $N$ and hence $NK=NL$ thuss we are done .

Without loss of generality, assume $AC>AB$. Let $OL\cap MB = D$ and $OK\cap MC = E$. Because $ODME$ is a parallelogram, we have $180 - \angle A = \angle DME = \angle DOE = \angle KOL \implies AKOL$ is cyclic. Let $NA\cap (AKOL) = F$. We will show that $\triangle NFK \cong \triangle NOL$ which will imply the result. Since $AN$ and $AO$ are isogonal, we have that $\angle KAF = \angle OAL \implies KF = OL$. Now note that $\angle AFO = \angle AKO = \angle ABM = \frac{\angle A}{2} + \angle B$ so $\angle NFO = \frac{\angle A}{2} + \angle C$. Also $\angle ONF = \angle OAN = (90 - \angle C) - (90 - \angle B) = \angle B - \angle C$. So $\angle FON = 180 - (\angle B - \angle C) - (\frac{\angle A}{2} + \angle C) = \frac{\angle A}{2} + \angle C = \angle NFO \implies NF = NO$. Also $\angle NOL = 360 - (180 - \angle A) - (\frac{\angle A}{2} + \angle C) - (90 - \angle B) = 90 + \angle B + \frac{\angle A}{2} - \angle C$. And, $\angle NFK = \angle AKF + 90 - \angle B = \angle B + \frac{\angle A}{2} + \angle B - \angle C + 90 - \angle B = \angle NOL$, implying the result.

We will use complex numbers with $(ABC)$ as the unit circle and $OM$ as the real line. Let $S$ be the antipode of $M$ with respect to the unit circle. Set $O=0,A=a,B=b,C=c,M=-1,S=1$. We have the special property that $bc=1$ since $b$ and $c$ are each others' conjugates as they are reflections across the real axis. Clearly $N$ is the reflection of $A$ across the imaginary axis, so by Lemma 7, we have $n=i+(-i)-i(-i)\overline{a}=-\frac 1a$. Now we compute $k$ and $l$. Let $U$ be the midpoint of $\overline{BS}$ and note that $u=\frac{b+1}2$. Observe that $OU\parallel MB$ as $\frac{u-o}{b-m}=\frac{\frac{b+1}2}{b+1}=\frac12$ is clearly real. So we can compute $K$ as $\overline{OU}\cap \overline{AB}$. By Proposition 6,\begin{align*} k&=\frac{(\overline{a}b-a\overline{b})(u-o)-(a-b)(\overline{u}o-u\overline{o})}{(\overline{a}-\overline{b})(u-o)-(a-b)(\overline{u}-\overline{o})}\\&=\frac{\left(\frac{b^2-a^2}{ab}\right)\left(\frac{b+1}{2}\right)-0}{\left(\frac{b-a}{ab}\right)\left(\frac{b+1}{2}\right)-(a-b)\left(\frac{b+1}{2b}\right)}\\&=\frac{\left(\frac{b^2-a^2}{ab}\right)\left(\frac{b+1}{2}\right)-0}{\left(\frac{b-a}{ab}\right)\left(\frac{b+1}{2}\right)+\frac{a(b-a)}{a}\left(\frac{b+1}{2b}\right)}\\&=\frac{(b^2-a^2)(b+1)}{(1+a)(b-a)(b+1)}\\&=\frac{a+b}{a+1}\end{align*}and similarly $l=\frac{a+c}{a+1}$. Observe that\begin{align*} |NK|^2&=(k-n)\overline{(k-n)}\\ &= \left( \frac{a+b}{a+1}-\left(-\frac1a\right)\right)\left(\frac{\frac 1a+\frac 1b}{\frac 1a+1}-(-a)\right)\\&=\left( \frac{a^2+ab+a+1}{a(a+1)}\right)\left(\frac{\frac{a+b}{ab}}{\frac{a+1}{a}\cdot\frac{b}{b}}+a\right)\\&=\left( \frac{a^2+ab+a+1}{a(a+1)}\right)\left(\frac{a^2b+ab+a+b}{b(a+1)}\right)\end{align*}and similarly$$|NL|^2=\left( \frac{a^2+ac+a+1}{a(a+1)}\right)\left(\frac{a^2c+ac+a+c}{c(a+1)}\right).$$So it suffices to show that$$c(a^2+ab+a+1)(a^2b+ab+a+b)=b(a^2+ac+a+1)(a^2c+ac+a+c).$$But since $bc=1$, \begin{align*}c(a^2+ab+a+1)(a^2b+ab+a+b)&=b(a^2+ac+a+1)(a^2c+ac+a+c)\\ (ca^2+abc+ac+c)(a^2b+ab+a+b)&=(a^2b+abc+ab+b)(a^2c+ac+a+c)\\(ca^2+a+ac+c)(a^2b+ab+a+b)&=(a^2b+a+ab+b)(a^2c+ac+a+c)\end{align*}which is clearly true, so we are done.

Let D be the midpoint of arc BAC in circumcircle. first we show that LC = KB. Let K' and L' be points on MB and MC such that ∠KK'B = ∠LL'C = 90. It's easy to show that KK'B and LL'C are congruent. ∠DOL = ∠DMC = ∠LAD so ADLO is cyclic. ∠LOK+∠A = ∠CMB+∠A = 180 so ALOK is cyclic. So ADLOK is cyclic. ∠DKA = ∠DOA = 2∠DBA so DK = KB and similarly DL = LC so DK=KB=LC=DL. Now lets show DKN and DLN are congruent. DK = KL , ∠KDN = ∠BDM = ∠CDM = ∠LDN , DN = DN ---> DKN and DLN are congruent ---> KN = LN