My solution: We have the lemma Lemma: Given triangle $ABC$, $AD,BE,CF$ are altitudes of triangle and $H$ is orthocenter. Let $X,Y$ on $AB,AC$ such that $XY \parallel EF$. $K,L$ are projections of $H$ onto $CX,BY$. Then we have $D,K,L,M$ are cyclic. Prove: Let $P,Q,R,S$ be the midpoints of $BY,CX,AB,AC$. We have $HKDC,HKXF$ are cyclic, i.e $BXKD$ is cyclic and $MQ || BX$ so $DKQM$ is cyclic.(1) Similarly we have $MDPL$ is cyclic. (2) We have $DMRS$ is cyclic (Euler's circle), and $\frac{SQ}{RP}=\frac{AX}{AY}=\frac{AC}{AB}=\frac{DS}{DR}$ So $\triangle DQS \sim \triangle DPS \rightarrow DMPQ$ is cyclic (2) From (1), (2), (3) we have $D,M,P,Q,K,L$ is cyclic. Done Now we back to main problem.

Attachments:

parmenides51 wrote: The altitudes of the acute-angled non-isosceles triangle $ ABC $ intersect at the point $ H $. On the segment $ C_1H $, where $ CC_1 $ is the altitude of the triangle, the point $ K $ is marked. Points $ L $ and $ M $ are the bases of perpendiculars from point $ K $ to straight lines $ AC $ and $ BC $, respectively. The lines $ AM $ and $ BL $ intersect at $ N $. Prove that $ \angle ANK = \angle HNL $. Let $BB_1,CC_1$ are altitude. $ML,B_1C_1$ cuts $BC$ at $R,S$. We know that $BMLA$ is cyclic and let $O$ be the center then we have $K,N,O$ are conlliner Let $P,Q$ are projection of $H$ onto $BL,AM$. We have $\angle HNL=\angle HQP$ and $HQ \perp AN$ so we just have prove $PQ \perp NO$. By Brocard then we have to prove $PQ \parallel CR$. $B_1X$ cuts $ML$ at $Z$. By chasing angle we easy to see that $M,Q,Z,B_1,A$ are cyclic. Let $(MQZ)$ cuts $(RQC_1)$ at $Y$ then $C,R,Y$ are collinear Now we have $XB^2=XC_1.XS$ and $RL || B_1S$ then chasing ratio we get $XC_1.XR=XZ.XB_1$ So we have $X,Y,Q$ are collinear. So we obtain: $\angle PQX=\angle PC_1X=\angle CLB=\angle CMA=\angle CYQ \Rightarrow PQ || CR$. We are done. P/s: i think this problem look very simple but it's not easy.

Attachments:

Let $AH \cap BC = A_1, BH \cap AC = B_1, MK \cap AB = F$ and $LK \cap AB = G$. Then $AB_1A_1B$ is cyclic and $\angle B_1AA_1 =\angle B_1BA_1$. Since $ALKC_1$ and $BMKC_1$ are cyclic, we have $CL\cdot CA = CK\cdot CC_1= CM\cdot CB$ and $ALMB$ is cyclic. Thus, \[\angle LAM = \angle LBM \implies \angle B_1AA_1 +\angle HAN = \angle B_1BA_1 + \angle HBN\]which gives $\angle HAN =\angle HBN$. Now, note that $AA_1 \parallel MF$ and $BB_1 \parallel LG$, triangles $AHB$ and $FKG$ are similar. Also, as $ALMB$ is cyclic, triangles $LNM$ and $ANB$ are similar and $LFGM$ is cyclic. Thus, triangles $LKM$ and $FKG$ are similar. This implies triangles $LKM$ and $AHB$ are similar and \[\frac{LN}{AN} = \frac{LM}{AB} = \frac{LK}{AH}.\] With $\angle NLK = \angle HBN = \angle NAH$, we obtain that triangles $NLK$ and $NAH$ are similar. This yields \[\angle LNK = \angle ANH \implies \angle BNK = 180^\circ - \angle LNK = 180^\circ - \angle ANH = \angle MHN\]so that $\angle HNB = \angle HNM + \angle MNB = \angle KNB + \angle MNB = \angle KNM$ and \[\angle ANK = 180^\circ - \angle KNM = 180^\circ - \angle HNB = \angle HNL.\]

It's obvious that ABML inscribed in (O), let (KC) cuts CG at J then J is Miquel point of ABML-CG but ON perpendicular to CG at Miquel point (Brocard) so O,N,K collinear I=(KNM)-CH, AI-(KNM)=Q then KIN=KMN=NAH so AHNI cyclic, notice that AQM=ANI=AHI=180-B so Q on (ABML). Let JB cuts (O) at Q so from OB^2=OQ^2=ON.OL so ONQB cyclic so QNK=OBQ=90-BAQ=90-QMC=KMQ so QKNM cyclic. AQ cuts CH at I then MQI=B=MKI so MKNI cylic hence ANH=AIH=QNK so ANK=HNL so q.e.d

It suffices to show that $\triangle MNK=\triangle BHN$ $CL \cdot CA=CK \cdot CC_1=CM \cdot CB \rightarrow ALMB$ is cyclic. Thus $\angle HAM=\angle NMK$ and $\angle HBN=\angle HAN$ then $\angle NMK=\angle HBN$ Now it suffices to show that $\frac{HB}{MK}=\frac{NB}{NM} \Longleftrightarrow \frac{NB}{NM}=\frac{AB}{LM}$ $LM$ in cyclic quadrilateral $= CK \cdot Sin (\angle C) \rightarrow \frac{AB}{LM}=\frac{c}{CK \cdot Sin (\angle C)}$ $\frac{HB}{MK}=\frac {\frac{1}{Sin(\angle C)} \cdot A_1B}{Cos(\angle B) \cdot CK}=\frac {\frac{1}{Sin(\angle C)} \cdot Cos(\angle B) \cdot c}{Cos(\angle B) \cdot CK}=\frac{c}{CK \cdot Sin (\angle C)}$ And we are done.

Attachments:

First notice that $\frac{CM}{CL}=\frac{cos \angle HCB}{cos \angle HCA}=\frac{CA}{CB}$. so$\triangle CLM$ is similar to $\triangle CBA$. And by simply angle chasing we have $\triangle NLM$ similar to $\triangle NBA$. So $\triangle CLM(N)$ is similar to $\triangle CBA(N)$ Then $\angle ANK=\pi -\angle MNK=\pi - \angle BNH=\angle LNH$. Done!