Given triangle $\triangle ABC$. Any circle with center $J$ passing through $BC$ intersects $AC$ and $AB$ at $E$ and $F$ respectively. Let $X$ be the point such that triangle $\triangle XBF$ is similar to $\triangle JCE$ ($X$, $C$ are on the same side of $AB$). Let $Y$ be the point such that triangle $\triangle YCE$ is similar to triangle $\triangle JBF$ ($Y$, $B$ are on the same side of $AC$). Prove that $XY$ always passes through the orthocenter of triangle $\triangle ABC$.
I'm preparing for maths exam in high school, but I can't find out any clue, please help me, I appreciate any help!
Before i started any thoughts in the direction of solving the question, i was needing a way to construct the figure. My figure looks low with a lot of unneeded and a few needed points as follows:
Now i need some time to explain the notations, and how $X, Y$ were constructed. Solution starts later, when the actors have their clear roles.
We start with $\Delta ABC$ with orthocenter $H$, and let $J$ be a point on the perpendicular bisector $\alpha$ of $BC$. Let us denote by $z$ the (signed) measure of the angles in $B,C$ of $\Delta JBC$ (which is isosceles). The points $E\in AC$ and $F\in AB$ are chosen so that $E\ne C$, $F\ne B$, and the circle centered in $J$ with radius $JB=JC$ passes through $E,F$. Let us use $2x$, $2y$ for the measures of the angles in $J$ in $\Delta JCE$ and $\Delta JBF$. A prime decoration of a point $Z$ denotes the point $Z'$ which is the projection of $Z$ on $AB$. A double prime (second) decoration of a point $Z$ denotes the point $Z''$ which is the projection of $Z$ on $AC$. From now on we will no longer need $E,F$. More relevant are the points $J',J''$. We have $x=\widehat{J''JC}$, $y=\widehat{J'JB}$. Working in $\Delta BJJ'$ and $\Delta CJJ''$ we see that $x,y$ are in relation to $z$, and to the data of $\Delta ABC$: $$ \begin{aligned} x &= 90^\circ-\widehat{JCJ''}=90^\circ-(C+z)\ ,\\ y &= 90^\circ-\widehat{JBJ'}=90^\circ-(B+z)\ ,\\ \end{aligned} $$ We will use for this triangle the conventional notations, $A,B,C$ also for the angles (instead of $\hat A, \hat B, \hat C$), $a,b,c$ for the sides (lengths), $h_a,h_b,h_c$ for the heights (lengths), $R$ for the circumradius. Let also $R_J$ denote the length $R_J=JB=JC=\frac a2\cos z$.
The points $X,Y$ were constructed as follows (to have the OP property). Let $E_1$ be the reflection of $E$ w.r.t. the perpendicular bisector $\alpha$. We draw a parallel to $JE_1$ through $F$, it intersects $BJ$ in a yellow point, take it together with $B,F$, draw the circle through these three points, and intersect this circle with the angle bisector $JJ'$ of $\widehat{BJF}$. There are two intersection points, we make the choice as in the figure, the yellow bulled is denoted by $X$ in the figure. Then the angle $2x$ from $\widehat {CJE}=\widehat {BJE_1}$ is thus transferred through the first yellow point to the second yellow point $X$. A similar construction was done for $Y$.
So far we have only a valid picture in geogebra. And geogebra confirms optically the colinearity of $H,X,Y$. What about a proof?
The proof i can type is unfortunately analytic. Well, one can use vectors instead, but we are not working in a synthetic manner. We break, organize the plane w.r.t. two directions, $JJ'$ and $JJ''$. Each point $Z$ has two coordinates w.r.t. these axes, we build parallels to the named axes $JJ''$ and $JJ'$, each parallel intersects the "other" axis in a point, and we read off the coordinates $(Z_{JJ'}, Z_{JJ''})$ obtained in this manner. For instance, $J=(0,0)$, and $X=(JX,0)$, and $Y=(0,JY)$. (The sign must be taken carefully in $JX,JY$, we use oriented axes $JJ'$ and $JJ''$.)
It remains to compute all coordinates of the points $H,X,Y$ explicitly, and verify the colinearity, which translates algebraically as a linear (affine) dependency. The point $H$ is the more complicated ingredient. Which is the component of $H$ on $JJ'$? We can read it on $JJ'$, but we can also read it off on any parallel to this line. The parallel through $J''$ is such a line, it is the one first height in $\Delta BJ''A$, and we intersect it with a second height $BH''\|JJ''$, and measure the part of this first height between the parallels $BH''\|JJ''$. So we need the distance between $J''$ and the orthocenter $\Phi:=H_{\Delta BJ''A}$ of the triangle in the index. I need a new picture for this isolated situation:
The component of $H$ on $JJ'$ is the one orange segment, it is congruent to the other orange segment $J''\Phi$, so it is $$ \begin{aligned} \Phi J'' &=\frac{H''J''}{\cos \Phi J''H''}=\frac{H''J''}{\sin A}\ ,\\ \Phi J''\sin A &= H'' J''= BJ\cos\underbrace{\sphericalangle(BJ,H''J'')}_{y+(90^\circ-A)} = R_J\cos (C-z) \ . \end{aligned} $$ (Above $y$ is seen as $\widehat{BJJ'}$ and $(90^\circ-A)$ as $\widehat{\Phi J''H''}$.) The other component is easily transposed in a formula.
So let us compute in detail. In $\Delta XBJ$ the sine theorem gives $JX:JB$ as $\sin\widehat {JBX} :\sin \widehat {JXB}$. So $$ \begin{aligned} JX &=R_J\frac {\sin(B-C)}{\cos(C+z)}\ , \qquad\text{ and similarly }\\ JY &= R_J\frac {\sin(C-B)}{\cos(B+z)}\ .\text{ So we have the needed components:} \\[2mm] H &=\frac 1{\sin A}(H''J''\ ,\ H'J') =\frac {R_J}{\sin A}(\ \cos (C-z)\ ,\ \cos (B-z)\ )\ ,\\ X &=(JX, 0)=R_J\left( \ \frac {\sin(B-C)}{\cos(C+z)}\ ,\ 0\ \right)\ ,\\ Y &=(0, JY)=R_J\left( \ 0\ ,\ \frac {\sin(C-B)}{\cos(B+z)}\ \right)\ . \end{aligned} $$ It remains to compute the following determinant, we will use $\sim$ to mark an equality up to some known factor. $$ \begin{aligned} & \begin{vmatrix} 1 & H''J''/\sin A & H'J'/\sin A\\ 1 & JX & 0 \\ 1 & 0 & JY \end{vmatrix} \\ &\qquad\sim \begin{vmatrix} \sin A & \cos(C-z) & \cos(B-z)\\ \cos(C+z) & \sin(B-C) & 0 \\ \cos(B+z) & 0 & \sin(C-B) \end{vmatrix} \\ &\qquad\sim -\sin A\sin^2(B-C) +\sin(B-C)(\cos(C-z)\cos(C+z)-\cos(B-z)\cos(B+z)) \\ &\qquad\sim -2\sin A\sin(B-C) +\Big((\cos 2C + \cos 2z)-(\cos 2B+\cos 2z)\Big) \\ &\qquad\sim -2\sin (B+C)\sin(B-C) +(\cos 2C -\cos 2B) \\ &\qquad=0\ . \end{aligned} $$ The determinant vanishes, so the points $H,X,Y$ are colinear.
$\square$