See the diagram as the definition of the point. Notice that $\triangle DXY \sim \triangle ABC$. So let $ABC$ be the reference triangle. Notice that $DXY$ has an orthic structure, so all lengths can be trig-bashed. (Very easy bash) Finished by Menelaus.

Solved with Shreyasharma. We initially gave up and complex bashed when I realized that synthetic isn't actually too hard. We first make the following two important observations. Claim :$BX$ and $CY$ intersect on $(ABC)$. Proof : This is well known, due to the right angles these lines intersect at the $A-$antipode. It is also well known that $BHCA'$ is a parallelogram which will come in handy later. Claim : Lines $AA'$ and $KL$ are parallel. Proof : Simply notice that, \[\measuredangle YLK = \measuredangle YXK = \measuredangle AXB = \measuredangle 90 + \measuredangle XAB = \measuredangle XBA = 180 + \measuredangle CBA = 180 + \measuredangle CA'A = \measuredangle YA'A \]which proves the claim. Now, we let $H' = \overline{KL} \cap \overline{AH}$. Note that by the previously established parallel lines, \[\frac{AY}{H'Y}=\frac{YA'}{YL}\]Further, since $XL \perp CY$, it is easy to see that $XL \parallel AC$ as well. Thus, \[\frac{YX}{YA} = \frac{YL}{YC}\]Taking the product of these two relations, we conclude that, \begin{align*} \frac{YX}{YH'} &= \frac{YA'}{YC}\\ YH' &= \frac{YX\cdot YC}{YA'}\\ &= \frac{YC \cdot \sin{\angle YA'B}}{\sin \angle YXA'}\\ &= \frac{YC \cdot \sin \angle CAB}{\sin \angle ABC}\\ &= \frac{YH'}{YC} &= \frac{\sin \angle A}{\sin \angle B} \end{align*}since $\measuredangle YA'B = \measuredangle CAB$ and $\measuredangle A'XY = \measuredangle BXA = \measuredangle ABC$. But, since $BHCA'$ is a parallelogram, we can also see that, \[\frac{YH}{YC} = \frac{\sin \angle YCH }{\sin \angle YHC } = \frac{\sin \angle YA'B}{\sin \angle BXA} = \frac{\sin \angle A}{\sin \angle B}\]From which it is quite clear that we have $H'=H$ and we conclude that points $H,K$ and $L$ are collinear as was desired. Remark: It was interesting to note that several neat results followed from this proof. One of the things we came across was the fact that if $H_A$ is the intersection of $(ABC)$ and the $A-$altitude, then quadrilaterals $BH_AKH$ and $CH_ALH$ are cyclic. This is because, \[\measuredangle BKH = \measuredangle XYL = \measuredangle AYC = 90 + \measuredangle HAC = \measuredangle BCA = \measuredangle BH_AA\]is true once we establish the collinearity claimed in the problem.

It is just an angle chase. WLOG assume that $AB>AC$. Let $A_1$ be the antipode of $A$ and let $AX\cap(ABC)=T$. Then $\angle XA_1Y=\alpha$ and $\angle A_1XY=\angle BXA = \beta$. We also have $\angle YHC=\beta.$ Note that $A_1LTX$ ic cyclic because of $\angle A_1LX=\angle A_1TX=90^{\circ}$. From this it follows that $\angle A_1XT=\angle TLY=\beta.$ Thus, $\angle YHC =\beta=\angle TLY $, so quadrilateral $TLHC$ is cyclic as well. Since $CH=CT$ $(T$ is the reflection of $H$ wrt $BC)$, we have that $\angle HLC = \beta.$ However, $\angle A_1LK=\beta$, so we are done.

Pretty quick to complex bash if you don't screw up the first time Anyways, set $(ABC)$ as the unit circle. Let $D$ be the $A$-antipode. Note that $H_a$, the reflection of $H$ over $\overline{BC}$ has coordinates $-bc/a$ and $d = -a$. Now to find $X = \overline{BD} \cap \overline{AH_a}$, \begin{align*} x &= \frac{-ab(a - bc/a) + bc(-a + b)}{-ab + bc}\\ &= \frac{-a(a - bc/a) + c(-a + b)}{ - a + c}\\ &= \frac{a^2 - bc + ac - bc}{a - c}\\ &= \frac{a^2 + ac - 2bc}{a - c} \end{align*}Now the foot from $X$ to $\overline{CD}$ is given by, \begin{align*} k &= \frac{1}{2}\left( c - a + x + ac\overline{x}\right)\\ &= \frac{1}{2} \left(c - a+ \frac{a^2 + ac - 2bc}{a - c} + \frac{c/a + 1 - 2a/b}{1/a - 1/c} \right)\\ &= \frac{1}{2}\left(c - a + \frac{a^2 + ac - 2bc}{a - c} + \frac{bc^2 + abc - 2a^2c}{b(c - a)} \right)\\ &= \frac{1}{2b(c - a)}\left(b(c - a)^2 - b(a^2 + ac - 2bc) + (bc^2 + abc - 2a^2c) \right)\\ &= \frac{1}{2b(c - a)} \left(2bc^2 - 2abc + 2b^2c - 2a^2c \right)\\ &= \frac{bc(b + c) - ac(a + b)}{b(c - a)} \end{align*}Similarly we find that, \begin{align*} l = \frac{bc(b + c) - ab(a + c)}{c(b - a)} \end{align*}Now we check the collinearity. First shift by $a + b + c$ so that, \begin{align*} k &= \frac{bc(b + c) - ac(a + b) - b(c - a)(a + b + c)}{b(c - a)}\\ &= \frac{a (a + b) (b - c)}{b(c - a)} \end{align*}Also then we have, \begin{align*} l &= \frac{a(a + c)(c - b)}{c(b - a)} \end{align*}Scaling down be $a(b - c)$ we have, \begin{align*} k &= \frac{a+ b}{b(c - a)}\\ l &= \frac{a + c}{c( a- b)} \end{align*}Scaling up by $bc(a - b)(a - c)$ we have, \begin{align*} k &= -c(a - b)(a + b)\\ l &= b(a - c)(a + c) \end{align*}Expanding these things we have, \begin{align*} k &= -a^2c + b^2c\\ l &= a^2b - c^2b \end{align*}Then to check the collinearity with the origin it suffices to show that, \begin{align*} \frac{-a^2c + b^2c}{a^2b - c^2b} \in \mathbb{R} \end{align*}Taking the conjugate we have, \begin{align*} \frac{-1/a^2c + 1/b^2c}{1/a^2b - 1/c^2b} &= \frac{-b^2c + a^2c}{c^2b - a^2b}\\ &= \frac{-a^2c + b^2c}{a^2b - c^2b} \end{align*}as desired.

sol that is a cool enough to actually post: Let $M$ be the UNIQUE intersection of $KL$ and $AH$. Clearly, $M$ is the UNIQUE point on $AH$ that lies on the polar of $BX \cap CY$. We desire to show that $H$ is said point, that $H$ indeed lies on the polar of $BX \cap CY$. By La Hire, this is equivalent to showing that $BX \cap CY$ lies on the polar of $H$. Let $D$ be the projection of $BX \cap CY$ onto $AH$. Since $X,Y$ and the center of the circle all lie on $AH$, the polar of $H$ is perpendicular to $AH$. As a result, we just want to prove $D$ lies on the polar of $H$. Since $D,H$, and the center of the circle are collinear, this is equivalent to proving that $D,H$ are inverses around the center of the circle. Let $N$ be the center of the circle, then we are just trying to prove that $NX^2 = ND \cdot NH$, where lengths are directed to ensure $D$ and $H$ are actually collinear in the right order. Let $E$ be the foot from $A$ to $BC$. This is equivalent to $4(EX + EY - 2EX)^2 = 4(EX + EY - 2ED)(EX + EY +2EH) $ which is equivalent to $ED^2= EY \cdot EX $. This is pretty easy, we have $ED = 2R \cos B \cos C$, $EX,EY = 2R \frac{\sin B \cos C}{\sin C}$ and the reverse in some order, so we are done. woah nontrivial difference of squares usage

Obvious by harmonic bundles. Referring to KI_HG's diagram the problem is equivalent to showing $(H, F; Y, X)=-1$ which is just projecting the pencil $(DH, DF; DB, DC)$ onto line $AH$.

Let $Q= KX \cap LY$ and $ R=DQ \cap XY$ with $M$ as midpoint of $BC$. ($D=BX \cap CY$) Note $Q$ is orthocenter of $\triangle DXY$ hence $DR \perp XY \Rightarrow DR \parallel BC$ and $\measuredangle DRA = 90$ hence $R$ lie on $(ABC)$. $$(B,C,M,\infty{\overline{BC}})\stackrel{D}{=}(X,Y;H,F)-1$$ hence by harmonic property $\overline{L-K-H}$