Let $H$ be the orthocenter of $ABC$, and let $M$ be the midpoint of $\overline{KL}$. Notice that \[\angle DBK=\angle DMK=\angle DCL=\angle DML=90^\circ,\]so $BDMK$ and $CDML$ are cyclic. We have \[\angle ABM=\angle KDM=90^\circ-\angle DKM=90^\circ-\angle BAC=\angle ABH,\]and similarly, $\angle ACM=\angle ACH$. This means $M=H$, so $H$ lies on $\overline{KL}$, as desired.

Ok somehow this is harder than p6 by a little bit (where a little bit means this is 2 mins solve (well noob me ig)) Let $(KBD) \cap (DLC)=H$ now $\angle KHD+\angle LHD=\angle KBD+\angle LCD=180$ so $H$ lies in $KL$ but note the perpendicular angles so in fact $DK=DL$ and $DH \perp KL$ so $H$ is midpoint of $KL$ now $\angle KLD=180-2\angle BAC$ and from here using the cyclics we get $\angle ABH=\angle ACH=90-\angle BAC$ which tells that $H$ is ortocenter of $\triangle ABC$ thus we are done.

Fun fact: the phrasing of the problem hints that $AD$ is a symmedian in $AKL$ and hence $AH$ bisects $KL$, though this is certainly not needed to solve the problem.

Let $H$ be the orthocenter of $ABC$. We redefine $K$ as the second intersection between $AB$ and $(BDH)$ and $L$ as the second intersection between $AC$ and $(CDH)$. By Thales’, $$\measuredangle DHK = \measuredangle DBK = 90^{\circ} = \measuredangle DCL = \measuredangle DHL$$which means $H \in KL$. By the Orthocenter Reflection Lemma, we know $D$ and $H$ are symmetric about the midpoint of $BC$, meaning $BDCH$ is a parallelogram. Now, $$\measuredangle DKL = \measuredangle DKH = \measuredangle DBH = \measuredangle BDC = \measuredangle BAC = \measuredangle KAL$$so $DK$ is indeed tangent to $(AKL)$. Similarly, we deduce that $DL$ is tangent to $(AKL)$. Hence, we have recovered the original definitions of $K$ and $L$ respectively, which finishes. $\blacksquare$

@2above you should prove that there is only a single pair of $K,L$ which satisfy the problem hypothesis, since a priori the property of $DK$ and $DL$ being tangent to $(AKL)$ might be true for all pairs which are collinear with the orthocenter or something. Anyways here's a somewhat overkill solution: Let $X,Y$ be the midpoints of $AK,AL$ respectively, and $O$ the circumcenter of $\triangle ABC$. Since $\angle KDL = \pi - 2\angle CAB$ and by homothety $\angle XOY = \angle KDL$, we have $\angle XOY + \angle COB = \pi$, so $O$ has an isogonal conjugate in the quadrilateral $CBXY$ (which is equivalent to saying that the conjugate of $O$ in $\triangle ABC$ is the same as the conjugate of $O$ in $\triangle AXY$). The conjugate in $\triangle ABC$ is the orthocenter, while the conjugate in $\triangle AXY$ is the point $H$ such that $AXHY$ is a parallelogram. Since these points coincide, $H$ is the orthocenter and by homothety at $A$ it's also the midpoint of $KL$. $\square$

ARMO 2013

$AO$ is the $A$-symmedian of $AKL$, so $AH$ is median of $AKL$. Let $M$ be the midpoint of $KL$, then, $(P_\infty, M; K, L) = -1$. Let $P = HD \cap (ABC) \implies (AP, AH; AB, AC) = -1 = (P_\infty, M; K, L) \implies KL \parallel AP$. Let $l$ be the perpendicular bisector of $KL \therefore$ $l \parallel PD$, but $D \in l$ and $D \in PD \therefore PD=l$. So, $M = l \cap AH = PD \cap AH = H \implies H \in KL.$ $_\blacksquare$

Let $M$ be midpoint of $KL$. We actually show $M\equiv H$, the orthocentre of $ABC$. $\angle DMK = 90^{\circ}=DBK$ and hence $\square DMKB$ is cyclic. Hence, $\angle DBM = \angle DKM = \angle KAL = \angle BAC = \angle DBH$. Hence, $H$ lies on $BM$. Similarly, $H$ also lies on $CM$. Hence, $M\equiv H$, and we are done!

Let $M$ be the midpoint of $KL$ .Then angle chase using the properties of symmedian and thats it.

Man, how rusty have I gotten in geo for not doing geometry for so long. Firstly, from the definition of the points $K$ and $L$ it is clear that the point $D$ becomes the intersections of the tangents to $\odot(AKL)$ at $K$ and $L$ and thus $AD$ becomes $A-$symmedian. Now $H$ denote the midpoint of segment $KL$. Thus we also have that $AH$ is the $A-$median and thus $AD$ and $AH$ becomes isogonals, that is the orthocenter of $\triangle ABC$ lies on $AH$. Now $DK$ and $DL$ being tangents, we have $DK=DL$ and thus $DH$ is the perpendicular bisector of $KL$. So we get $\measuredangle DHK=90^\circ=\measuredangle DBK$ which gives $DBKH$ is cyclic. And similarly $DCLH$ is also cyclic. To finish we have, \begin{align*} \measuredangle BHC&=\measuredangle HBD+\measuredangle BDC+\measuredangle DCH\\ &=\measuredangle BAC+\measuredangle HKD+\measuredangle DLH\\ &=\measuredangle BAC+\measuredangle LKD+\measuredangle DLK\\ &=\measuredangle BAC+2\measuredangle LAK\\ &=\measuredangle CAB .\end{align*} This gives that the orthocenter of $\triangle ABC$ lies on $\odot(BHC)$ which upon combining with the fact that the orthocenter lied on $AH$, we conclude that $H$ is indeed the orthocenter of $\triangle ABC$ and we are done.

Solution : Construction : Let $X$ be a point on $LM$ such that $XL=XK$ & $H$ be the orthocenter of $\triangle{ABC}$. Claim 1 : $BKXD$ is cyclic Proof : Observe that $$\angle{KXD}=\frac{\pi}{2}=\angle{KBD}$$. So Claim $1$ is proved . Easy Angle Chasing gives $$\angle{DBX}=\angle{DKX}=\angle{BAC}=\angle{HBD}$$. Claim 2 : $CLXD$ is cyclic Proof : Similar approach like Claim $1$ . Finish : So from both Claim $1$ & $2$ we get that $H$ lies on both lines $XB$ and $XC \Longrightarrow H=X$. Therefore $KL$ passes through the orthocenter of $\triangle{ABC}$ .

Kind of similar configuration to Singapore MO Open 2022 Q1 WLOG let $HB$ intersect $KL$ at $B'$ and $HC$ intersect $KL$ at $C'$ Firstly, note that $BHCD$ is a parallelogram $$\angle{CBD}=\angle{HBC}=90-b$$$$\angle{HCF}=\angle{HBD}=\angle{HBC}+\angle{CBD}=90-c+90-b=a$$ But note that due to the tangent condition, $\angle{LKD}=\angle{KLD}$ This means that $BKB'D$ and $CLC'D$ are concylic Which then implies that $\angle{KB'D}=\angle{DC'L}=90$ which means that $C'=B'=H$ Q.E.D

Sol:- We have $\angle KDL=180^\circ-2A$ and $\angle BDC=180^\circ-A$.Hence if we let reflection of $K$ across $B$ to be $K'$ and reflection of $L$ across $C$ to be $L'$ we have $K'-D-L'$ collinear.So we can observe that $(KLK'L')$ is cyclic with diameter $K'L'$. So, $K'L \perp AC$ and $L'K \perp AB$. Hence $B-$ altitude and $C-$ altitude bisects $KL \implies$ orthocenter of $ABC$ is midpoint of $KL$.

Why was my sol deleted?? Anyway I'll leave a sketch, DEFINE $H$ to be the midpoint of $KL$,note that $DBKH,DCKH$ are cyclic.Now angle chase to find out that $DBHC$ is a parallelogram,since $H$ lies inside $\triangle ABC$, $H$ is the orthocentre!

Assume that H is the midpoint of the KL. Then we have AD is A-symmedian of the triangle AKL, therefore AH is isogonal of AD in KAL, so AH is an altitude of triangle ABC. Now the quadrilateral BKHD is cyclic, hence \angle HBD=\angle HKD=\angle BAC=\pi-\angle BDC. Therefore lines BH and CD are parallel, hence BH is an altitude of triangle ABC. We are done.

Let the altitude from $A$ cut $KL$ at a point $H$. By construction we have that line $AD$ (i.e. line $AO$) is a symmedian in triangle $AKL$, so it's isogonal to the $A$-median. Combining this with the well-known fact that $AH$ and $AO$ are isogonal in $ABC$, we get that $H$ must be the midpoint of segment $KL$. It then follows that $DH\perp KL$ by the tangencies. Now let line $DH$ cut $(ABC)$ again at a point $X$, so that $AX\perp DH\perp KL$. Projecting the harmonic bundle $(K;L;H;\infty_{KL})$ from $A$ onto $(ABC)$ then gives that points $B,C,X$ and $(AH)\cap (ABC)$ are harmonic, so $X$ is actually the $A$-queue point. The conclusion follows since this point is the intersection of the line passing through $D,M$ and $H'$ where $M$ is the midpoint of $BC$ and $H'$ the orthocenter (and $H$ is also on the $A$-altitude). $\square$

whoops Let $E$ and $F$ be the feet of the $B$- and $C$-altitudes respectively, $H$ be the orthocenter of $\triangle ABC$, and let $H_1=\overline{KL} \cap \overline{BE}$ and $H_2=\overline{KL} \cap \overline{CF}$. It is well-known that $BHCD$ is a parallelogram, so $\measuredangle DBH=\measuredangle HCD=\measuredangle FHE=\measuredangle BAC$. But we also have $\measuredangle DKL=\measuredangle KLD=\measuredangle ABC$, hence $DBKH_1$ and $DCLH_2$ are cyclic. From this, we find that $$\measuredangle KH_2D+\measuredangle DH_1L=\measuredangle KBD+\measuredangle LCD=0,$$hence $\measuredangle H_2DH_1=0$ and $H_1=H_2$, since they both lie on $\overline{KL}$. But then this means that $H_1$ lies on both $\overline{BE}$ and $\overline{CF}$, hence it equals $H$ and we have $H \in \overline{KL}$ as desired. $\blacksquare$

note that $AD$ is the symmedian line of $AKL$ let $H$ is the midpoint of $KL$ $\angle AHL + \angle DBL = 180^\circ$ so $DBLH$ is cyclic so $\angle A = \angle HLD = \angle HBD $ $\rightarrow \angle CBH = 90^\circ -\angle C $ $\rightarrow BE $ is altitude of $\Delta ABC$ $Q.E.D$

FHXB and FECX is a quad FC and EX are the diagonals of FECX FX and HB are the diagonals of FHXB AC1+C1E=AE AC1+C1C=AC Therefore,C1 and C intersect at E There is a triangle BEX Triangle BB1X and BEX is similar Therefore,C1,C,B,B1 are concurrent. Sorry for mistake,it was a solution for a elmo problems,I made mistake with ELMO and EGMO. Please don't consider that mistake.

Too easy. Let perpendicular from $A$ to $BC$ intersects $KL$ at poin $T$. Since $H$ lies on $AT$ and $O$ lies on $AD$, and from $O$ and $H$ isogonal, we have $AD$ and $AT$ are isogonal, since $AD$ is symmedian, we have $T$ is midpoint of $KL$. Then $DT$ is perp. to $KL$, then $(L T D C)$ is cyclic, then $\angle ACT=90-\angle TCD=90-\angle TLD=90-\angle BAC$, which means that $AT$ is perp. to $AB$ then $T$ is orthocenter of $ABC$, which completes the solution.

let $M$ the midpoint of $KL$. claim $1$: $M \in AH$ proof: $AD$ is $A-symmedian$ of $AKL$ then $\angle CAH= \angle KAD = \angle MAC$.$ _\blacksquare$ claim $2$: $LCDM$ cyclic proof: $\angle LCD= \angle ACD = \angle LMD. _\blacksquare$ claim $3$: $M \in CH$ proof: $180^{\circ}-\angle 2A = \angle KDL$$ \implies \angle LDM = 90^{\circ}-\angle A = \angle LCM = \angle ACM ._\blacksquare$ by claim $1$ and $3$ we have $ M = CH \cap AH =H$. and we are done.

Too Easy This problem is basically this https://artofproblemsolving.com/community/q2h535000p29584904

We do $\sqrt {bc}$ inversion centred at $A$. Let $H$ be the orthocentre of $\triangle ABC$. Let $E,F,G$ be the altitudes of $A,B,C$ onto $BC,AC,AB$ respectively. The inversion sends $E$ to $D$, and sends $F$ to the point on $AB$ such that $ACF'$ is right, and similar for $G'$. Now the orthocentre is the intersection of $(AFG)$ with $AE$, so $H'$ is the intersection of $F'G'$ with $AD$. It is trivial to see that $D$ is the orthocenter of $\triangle AF'G'$, so $AD$ is simply the altitude of $A$ onto $F'G'$. Now see where $K,L$ are sent to. $K'L'$ is tangent to $AK'E$ and $AL'E$ at $K',L'$. This is equivalent to $E$ being the $A-HM$ point of $AK'L'$. Now since $AE$ and $AD$ are isogonal, the orthocentre of $AK'L'$ is the intersection of the perpendicular to $AE$ through $E$ with $AD$, which is precisely the intersection of $BC$ with $AD$. We want to show that $K, H,L$ are collinear or that $A,K',H',L'$ are concyclic. Let $P$ be the intersection of $BC$ with $AD$, since $P$ is the orthocentre of $AK'L'$, $L'P$ is parallel to $CD$, and $K'P$ is parallel to $BD$. We claim that $L'K'$ is parallel to $F'G'$, but this is trivial since: $\frac{AL'}{AF'}=\frac{AP}{AD}=\frac{AK'}{AG'}$. Now we want the dilation at $A$ with power $\frac{AP}{AD}$ to take the circumcircle of $\triangle AF'G'$ to the "circumcircle" of $AK'H'L'$, hence it suffices to show that $H'$ is sent to a point on the circumcircle of $AF'G'$ in this dilation. Let $AD$ intersect $(AF'G')$ at a point $Q$. It suffices to prove that $\frac{AP}{AD}=\frac{AH'}{AQ}$ or just $AP\times AQ=AD\times AH'$. Now, $AP\times AQ=AC\times AG'=AD\times AH'$ by repeated Power of a Point, so we are done, as $AK'H'L'$ is cyclic and $KHL$ are collinear.

Define $M$ as the midpoint of $KL$ to form cyclic quadrilaterals $BKMD$ and $CLMD$. Proving $M=H$ boils down to showing $BDCM$ is a parallelogram, which is true as \[\angle CMB = \angle DKB + \angle CLD = \angle ALK + \angle AKL = 180-\angle A = \angle BDC\]\[\angle MBD = \angle MKD = \angle DLM = \angle DCM. \quad \blacksquare\]

xoink

Let $M$ be the midpoint of $KL$. The motivation behind considering the midpoint is that $AD$ is $A-$symmedian of $\triangle AKL$ so $AM$ will be the isogonal conjugate and by angle chasing we will have $\measuredangle CAM=\measuredangle CAH$) To Prove: $M$ is the orthocenter of $\triangle ABC$. Claim: $\odot(DBKM)$ Proof: $$\measuredangle DBK = \measuredangle DMK = 90^\circ$$So we have $$\measuredangle DBM = \measuredangle DKM = \measuredangle BAC = \measuredangle HBD$$and therefore $BM \perp AC$. Similarly $CM \perp AB$, so $M$ is the orthocenter of $ABC$.

From the tangents we have that $\angle DLK = \angle DKL = \angle CAK = \alpha$. We will prove that $H \equiv M$, where M is the midpoint of LK, then $H \in LK$. We know that M is the midpoint of LK $\Rightarrow$ $\angle LMD = \angle DMK = 90^{\circ}$ $\Rightarrow$ $\angle LDM = \angle KDM = 90 - \alpha$. Now $\angle LCD + \angle LMD = 90 + 90 = 180^{\circ}$ $\Rightarrow$ LMDC is cyclic. Similarly since $\angle KBD + \angle KMD = 90 + 90 = 180^{\circ}$, MKBD is cyclic. From the cyclic quadrilaterals it follows that $\angle LCM = \angle LDM = 90 - \alpha$ and $\angle KBM = \angle KDM = 90 - \alpha$ $\Rightarrow$ $\angle ACM = 90 - \alpha$ and $\angle ABM = 90 - \alpha$ $\Rightarrow$ $M \in C_h$ and $M \in B_h$, where $B_h$ and $C_h$ are the heights from B and C respectively $\Rightarrow$ $M \equiv H$ $\Rightarrow$ H is the midpoint of LK $\Rightarrow$ $H \in LK$ and we are ready.

Very beautiful problem! [asy][asy] import geometry; unitsize(4cm); pair A = dir(110); pair B = dir(220); pair C = dir(320); pair H = A+B+C; pair G = dir(250); pair M = (B+C)/2; pair O = 0; pair Y = reflect( perpendicular(O, line(G,M))) * G; pair D = -A; pair T = intersectionpoint(perpendicular(B,line(O,B)), perpendicular(C,line(O,C))); pair K = intersectionpoint(perpendicular(H,line(H,D)), line(A,B)); pair L = intersectionpoint(perpendicular(H,line(H,D)), line(A,C)); draw(A--B--C--cycle); draw(circle(A,K,L), dashed); draw(D--K); draw(D--L); draw(circle(A,B,C)); draw(Y--T); draw(A--G); draw(Y--B); draw(Y--C); draw(Y--G); draw(A--D); draw(K--L); draw(T--B); draw(T--C); draw(A--H--D--cycle, red+1.1); draw(Y--M--T--cycle, red+1.1); filldraw(A--H--D--cycle, invisible, red+1.1); filldraw(Y--M--T--cycle, invisible, red+1.1); filldraw(A--K--L--cycle, invisible, blue+1.1); filldraw(Y--B--C--cycle, invisible, blue+1.1); dot("$A$", A, dir(110)); dot("$D$", D, dir(290)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$H$", H, dir(0)); dot("$G$", G, dir(250)); dot("$M$", M, dir(135)); dot("$Y$", Y, dir(60)); dot("$T$", T, dir(270)); dot("$O$", O, dir(120)); dot("$K$", K, dir(180)); dot("$L$", L, dir(30)); [/asy][/asy] Notice that $AD$ is the $A$-symmedian in $\triangle AKL$. Thus \[ \frac{AK}{AL} = \frac{\sin KAD}{\sin LAD} = \frac{\cos C}{\cos B} = \frac{HC}{HB} \]where $H$ is the orthocenter of $\triangle ABC$. Define $G$ as the reflection of $H$ over $BC$ and $M$ as the midpoint of $BC$. Let $GM$ intersect $(ABC)$ again at $Y$. Claim: We have $\triangle YBC \sim \triangle AKL$. Proof. Then $YD$ and $YG$ are isogonal wrt $\angle BYC$ so $YBDC$ is harmonic, implying \[ \frac{YB}{BC} = \frac{DB}{DC} = \frac{GC}{GB} = \frac{HC}{HB} \]hence $\triangle YBC \sim \triangle AKL$. $\blacksquare$ Now, let $T$ be the pole of $BC$ wrt $(ABC)$. Then the ratio of the spiral similarity sending $\triangle YBC$ to $\triangle AKL$ is $\frac{YT}{AD}$ (both are the segments on the $A$-symmedian joining the vertex to the opposite pole). Moreover, $YDT$ is collinear by the definition of the symmedian. Claim: We have $\triangle MYT \sim \triangle HAD$. Proof. Note $\angle MYT = \angle GYD = \angle HAD$. If $O$ is the center of $(ABC)$, note that $OM \cdot OT = OY^2$ by the definition of pole. Thus $\triangle OMY \sim \triangle OYT$, and so \[ \frac{YM}{YT} = \frac{OM}{OY} = \frac{OM}{OB} = \cos A = \frac{AH}{AD} \]thus $\triangle MYT \sim \triangle HAD$. $\blacksquare$ But then the spiral similarity sending $YT$ to $AD$ (which is the same one that sends $\triangle YBC$ to $\triangle AKL$) also sends $YM$ to $AH$. By similarity then $H$ is the midpoint of $KL$. $\square$

Main part is just constructing the midpoint of $KL$(Let's say it $M$) . Then from angle chasing $H \equiv M$ (orthocenter lemma helps for that problem)

We note that AH perpendicular to BC and H is on KL.We want to prove H is the orthocenter.We get with AD is symmedian AH is median.We get DH perpendicular to KL.Then BKHD,LHDC is cyclic.We get with easy angle-chasing BH is perpendicular to AC.We are done.

Let $M$ be the midpoint of $KL$. Notice that $\angle DHK=\angle DBK=90^\circ$ so $BMDK$ is cyclic. Analogously we get $HLCD$ cyclic, which implies $$\angle HBD=\angle HCD = \alpha.$$However, we also have $\angle HBD=\angle HCD= \alpha$ so indeed $M=H$.