Let $O$ be the circumcentre of $ABC$, $O'$ be the circumcentre of $AKL$, $D$ be $AO \cap BC$. Let $AO \cap \odot AKL = A'$. Note, $\angle LKA' = \angle ABC$ so $KA' \parallel AB$ and similarly $LA' \parallel AC$ so a homothety wrt $D$ takes $A'KL \mapsto ABC$. So it is internal centre of similitude i.e. $(A, D; O, O') = -1$. If $M$ is the midpoint of $KL$ then $(MA, MD, MO, MO') \cap AH$ gives the conclusion. Just for the sake of completion, $A$ is the external centre of similitude as $AKL, ABC$ have the same angle bisectors at $A$, so the isogonal of the common altitude means $A, O, O'$ are collinear.

Let $AN$ be the diameter of $(ABC)$, $H$ is the feet of altitude from $A$, reflect $A$ over the midpoint of $LK$ to obtain $A'$, hence we just have to prove $H,N,A'$ are collinear. It's clear that $BCN,LKA'$ are homothetic. Since $HC*LH=AH^2=BH*HK\Rightarrow \frac {HC}{HK}=\frac {BH}{LH}$, therefore $H$ is the center of homothetic and the collinearity follows.

Consider midpoints of AK and AC,respectively P and R.Now,because M,P and R are colinear(MP//MR//BC),and OR//AL//NP,it will be enough to prove that MNP and MOR are homothetic.Now,consider the diameter AA'.Because triangles ACA' and ADL are similar,and ADC and ADK,that yelds that DK/AL=DC/CA',which is the end when we divide by two

junioragd wrote: Consider midpoints of AL and AC,respectively P and R.Now,because M,P and R are colinear(MP//MR//BC),and OR//MP//AK,it will be enough to prove that MNP and MOR are homothetic.Now,consider the diameter AA'.Because triangles ACA' and ADL are similar,and ADC and ADK,that yelds that DL/AK=DC/CA',which is the end when we divide by two Whats $N,M$ and i think $OR$ is paralllel to $AK$ is not correct

There is typo.Just change K and L

M is midpoint of the height from A,and N is the midpoint of KL

I fixed the typos,now it should be ok.

I cant see how $OR||MP||AL$ , $MP$ is wrong and where is $D$? Could you explain?

I fixed it again,it is OR//NP//AL and D is the foot of the height from A to BC.

Junioragd, perfect solution with messy bad writing. You are like a dirty gold . At first look like a mud , but in the end i understand your value and this solution.

Lemma: in $\triangle ABC$ let $E,F$ on $AB,AC$ and $G,H$ on $BC$ such that quadrilateral $EFGH$ is rectangle let $P$ be the center of this rectangle then the locus of point $P$ is the line $MN$ where $M,N$ are midpoints of $BC$ and altitude $AD$($D$ lies on $BC$). Prove:First assume that $EFGH$ is rectangle and $P$ is its center we prove that $P,M,N$ are collinear. Let $MB\cap EH=S$ because $EH||AD,AM=MD\longrightarrow ES=SH$ so $PS||BC$ similarly $AN$ bisects $EF$ at $T$ midpoint of $EF$ clearly $ET=PS$ note that $\frac{MS}{MB}=\frac{AE}{AB}=\frac{ET}{BN}=\frac{SP}{BN}$ so $N,P,M$ are collinear by thales theoream and the lemma proved. Back to the problem: Let $D$ foot of $A$ on $BC$ and $AL\cap \odot (\triangle ABC)=F,AK\cap \odot (\triangle ABC)=E$ because $\angle BAE=\angle CAF=90$ points $C,O,F$ and $B,O,E$ are collinear so $EFBS$ is a rectangle with center $O$ using the lemma we deduce that $M,N,O$ are collinear.($M,N$ midpoints of $KL,AD$) DONE

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My solution: $H$ is the projection of $A$ on $BC$, $M, N, O$ are the midpoints of $AH, KL$ and the circumcenter of $\triangle{ABC}$ Let $S$ be the antipode of $B$ WRT $(ABC)$ Obviously, $S\in{AK} (\because AK\perp{AB})$ and $CS\perp{CB}$ Now: $HC.HL = HA^2 = HB.HK$ $\Rightarrow \frac{HB}{HL} = \frac{HC}{HK}$ $\Rightarrow \frac{HB}{HL} = \frac{AS}{AK}$ $\Rightarrow$ The midpoints of $HA, BS, LK$ are collinear (ERIQ) $\Rightarrow \overline{M, O, N}$ Q.E.D

My solution : Let $ B^*, C^* $ be the antipode of $ B, C $ in $ \odot (ABC) $, respectively . Let $ O $ be the circumcenter of $ \triangle ABC $ and $ M $ be the midpoint of $ KL $ . Easy to see $ B^* \in AK $ and $ C^* \in AL $ . Since $ AM $ (A-median of $ \triangle AKL $) passes through the midpoint $ N $ of $ B^*C^* $ , so $ MO $ (M-median of $ \triangle MAH $) passes through the midpoint of $ AH $ ($ \because NO \parallel AH $) . Q.E.D

Complete the parallelograms $ALKX$ and $HBCA'$ where $H$ is the orthocenter of $\triangle ABC$ and so $A'$ is the antipode of $A$ in $\odot (ABC)$. By $\triangle ALK \mathrm{ \ homothetic \ to \ } \triangle HBC$ and $\triangle XKL \mathrm{ \ homothetic \ to \ } \triangle A'CB$, we have that the constructed parallelograms are homothetic, and the homothetic center is $AH \cap A'X \cap BC$ which is precisely $D$. By the homothety $\mathbb{H}(A, \frac{1}{2})$, we get our conclusion. PS.: Sorry to revive such an old discussion.

Let $M'$ the midpoint of the two intersections $(\neq A)$ of the $ABC$-circumcircle and $AK,AL$ say $K',L'$ . Let $P,M ,N$ the respective midpoints of $KL,BC$ and the altitude. By simple angle chase we deduce that $\angle K'AL' =\pi-\angle A$ and $AK,AL $ are isogonal WRT $\angle BAC $ so $\odot (ABC),\odot (AKL)$ are tangent at $A$ hence $K'L'\parallel BC $ besides $K'L'=BC $ implies that $O$ is their symmetry center i.e .$OM=OM'$ .Since $MM'\parallel AN$ then $PO$ cuts the altitude at its midpoint . RH HAS

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Let $M$ be the midpoint of $\overline{BC}$ and $B', C'$ be the $B$ and $C$-antipodes in $\odot(ABC)$; obviously $\overline{AC'L}$ and $\overline{AB'K}$ are collinear. If $M'$ is the reflection of $M$ over the circumcenter of $\odot(ABC)$, then it is also the midpoint of segment $\overline{B'C'}$ and so line $AM'$ bisects $\overline{KL}$, implying the conclusion.