Let $H$ be the orthocenter of a triangle $ABC$. Let $M$ be the midpoint of $BC$, and let $E,F$ be the feet of the $B$ and the $C$ altitudes onto the opposite sides. Let $X$ be the intersection of ray $MA$ with the circumcircle of $BHC$. Prove that $HX,EF$ and $BC$ concur at a point.
My attempt :
Let $HX$ and $BC$ intersect at point $T$. By the power of point we know that $TB.TC=TH.TX$, and so we need to prove that $TB.TC=TH.TX=TF.TE$ ; if $T,F,E$ are collinear then this is indeed true by the power of point, but how should I prove this collinearity? I wanted to use phantom point, but I don't know how. Any help is surely appreciated. Thanks!
The proof described below makes use of a special case of the three conics theorem shown in $\mathrm{Fig.\space 1}$. This theorem states that the radical lines of three circles are concurrent in a point known as their radical center (aka. power center). In the problem we are dealing with, the radical lines turn out to be the common chords, i.e. $AD$, $BE$, and $CF$, because each circle intersects with the other two.
To facilitate the methodology used, we need to introduce several geometrical entities, such as lines and circles, to the configuration described by OP in his/her problem statement. They are drawn in dotted line style so that OP can pick them out easily.
First, we deduce the statement given at the end of the problem description by assuming that the quadrilateral $DMXH$ (see $\mathrm{Fig.\space 2}$ is cyclic. Then, we give you a proof of the premise about the mentioned quadrilateral.
As shown in $\mathrm{Fig.\space 2}$, we construct the nine-point circle of $\triangle ABC$, which passes through the feet of the altitudes $E$, $F$, and $D$, and the midpoint $M$ of the side $BC$. We need to draw another circle, which goes through $A$, $F$, $H$, and $E$. This is possible because $ \measuredangle AFH = \measuredangle HEA = 90^o$.
With the introduction of these entities, $EF$ becomes the common chord of the nine-point circle $EFDM$ and the circumcircle of $BCEF$, while side $DM$ (i.e. $BC$) turns up as the common chord of the circumcircle of $DMXH$ and the nine-point circle $EFDM$. Since the quadrilateral $DMXH$ is cyclic, $$ \measuredangle HXA = \measuredangle HDM = 90^o.$$
This condition is necessary and sufficient for us to state that $X$ lies on the circumcircle of $AFHE$. Therefore, $HX$ is the common chord of the circumcircles $AFHE$ and $DMXH$.
Now, consider the circle triad containing the pair of circumcircles $AFHXE$ and $DMXH$, and the nine-point circle $EFDM$, of which each circle intersects with the other two. Therefore, they have three common chords, i.e. $HX$, $DM$, and $EF$ and they are concurrent according to the aforementioned three conics theorem.
The description given below outline how we prove that the quadrilateral $DMXH$ is cyclic. Pay attention to $\mathrm{Fig.\space 3}$.
$\underline{\mathrm{Lemma}}$
Point $H$ is the orthocenter of scalene triangle $ABC$, in which $M$ is the midpoint of the side $BC$. Points $D$, $E$, and $F$ are the feet of the $A$-, $B$-, and $C$-altitudes respectively. The line joining $A$ to $M$ intersects the circumcircle of the triangle $BCH$ at $X$. Prove that the quadrilateral $DMXH$ is cyclic
$\underline{\mathrm{Proof\space of\space the\space Lemma}}$
We denote the center of the circumcircle of $\triangle BCH$ as $O$. Join the orthocenter $H$ to $O$ and extend it to meet the circle at $V$. Construct the lines $HX$, $BV$ and $CV$. After denoting the midpoint of the segment $AH$ as $K$, draw a line joining $K$ and $O$. We mark the point of intersection between $KO$ and $HX$ as $U$. Our immediate aim is to prove that the line $AV$, once it is drawn, passes through $M$, the midpoint of $BC$.
The two angles, $ \measuredangle VCH$ and $ \measuredangle HBV$, are angles in semicircles. Since an angle in semicircle is a right angle, we have, $$\measuredangle VCH = \measuredangle HBV = 90^o. $$
Therefore, we shall write, $$ \measuredangle VCH = \measuredangle CFB\quad\text{and} \tag{1}$$ $$ \measuredangle HBV = \measuredangle CEB.\qquad\enspace \tag{2}$$
The two equalities (1) and (2) mean that $VC$ and $BV$ are parallel to $AB$ and $CA$ respectively. This pair of conditions is necessary and sufficient for us to state that $ABVC$ is a parallelogram. The two diagonals of a parallelogram bisect each other. Therefore, the diagonal $AV$ passes through $M$, the midpoint of $BC$.
Note that $HV$ is a diameter of the circle $BCH$, and, therefore, its center $O$ is the midpoint of $HV$. We have already marked the midpoint of $AH$ as $K$. Now, as given below, we can deduce a series of conclusions in succession.
$\quad 1.\space$ Since $O$ and $K$ are the midpoints of $HV$ and $AH$ respectively, $KO$ is parallel to $AM$.
$\quad 2.\space$ Since $KO$ is parallel to $AM$, $U$ is the midpoint of $HX$.
$\quad 3.\space$ Since $HX$ is a chord of the circle $BCH$ and $KO$ passes through $U$, $KO$ is perpendicular $\\$
$\qquad$ to $HX$.
$\quad 4.\space$ Since $KO$ is parallel to $AM$ and $KO$ is perpendicular to $HX$, $AM$ is perpendicular$\quad\space\enspace$ $\\$ $\qquad$ to $HX$.
The fourth deduction shows us that angle $MXH$ of the quadrilateral $DMXH$ is a right angle. We already know its opposite angle $HDM$ is also a right angle. Therefore, the quadrilateral $DMXH$ is quadrilateral.