Prove BC tangent to the circle (XYM)

Question

Given $\Delta BC$ inscribed $(O)$ and the altitudes $BE,CF$. Let $O'$ be the reflection of $O$ over line $BC$. Let $X,Y$ be the center of $(O'BE),(O'CF)$ respectively. Let $M$ be the midpoint of $BC$. Prove $BC$ tangents to the circle $(XYM)$

Here's what I've obtained so far: $MX \perp EB$ and $MY \perp CF$. Let $P,Q$ be the projections of $M$ on $EB,FC$ and let $H$ be the orthocenter, $D$ be the projection of $A$ on $BC$. We have $(H,P,Q,D,M)$ concyclic. Therefore, I'm thinking about using inversion on $H$ (which sent $(H,P,Q,D,M)$ to a line through $A$ and the intersection of $AM$ with $(AEF)$)

Answer 1

Proof

Let $H$ be the orthocenter of $\triangle ABC$.

Since $HB \cdot HE=HC \cdot HF$, $H$ lies on the radical axis of $\bigodot X$ and $\bigodot Y$. Hence, $XY \perp O~'H $. But $O~'H \parallel OA$, thus $XY \perp OA.$ Since $OA \perp EF$, therefore, $XY \parallel EF$.

Notice that $M$ is the center of $\bigodot BFEC$, and $BF$ is the radical axis of $\bigodot BFEC$ and $\bigodot X$. Hence, $MX \perp BE$. Similarily, $MY \perp CF$. Further, we may have that $MX \parallel AC, MY \parallel AB$.

Now, let's chase the angles. According to the parallel relations above, it's easy to have $$\angle XMB=\angle ACB=\angle AFE=\angle XYM.$$ Thus, $\angle XMB$ is necessarily the chord tangent angle of $\bigodot XYM$,namely, $BC$ is the tangent of that. We are done.

Answer 2

This is a beautiful problem. Here is first a proof involving trigonometry. (I will try to also give a synthetic proof, avoiding it.) But so far this is a solution, searching for a better one may be easier now.

First of all, let us show the following:

Lemma: Fix a triangle $\Delta ABC$. Let $M,N,P$ be the mid points of its sides, opposite to $A,B,C$ respectively. Let $(O)$ be the circumscribed circle, its center is $O$. Let $O'$ be the symmetric of $O$ w.r.t. $BC$ (and $M$). Let $(S)$ be the circle through $B,C,O'$, centered in $S$. Let $X,Y$ be the intersection with the side perpendicular bisectors of the sides $O'B$, and $O'C$. Then $$MX\cdot MP=MY\cdot MN\ .$$

Proof: The trigonometric proof is simple, since we know all angles in the two triangles $\Delta SMX$ and $\Delta SMY$. (So the bridge between $X$ and $Y$, the idea, is to consider $SM$ and relate $MX$, $MY$ to it.)

$$ \begin{aligned} \frac{MX}{MY} &= \frac{MX}{MS}\cdot\frac{MS}{MY} \\ &= \frac{\sin(90^\circ-x)}{\sin(180^\circ-y)} \cdot \frac{\sin(180^\circ-z)}{\sin(90^\circ-x)} = \frac{\sin z}{\sin y} \\ &=\frac{AB}{AC}=\frac{MN}{MP}\ . \end{aligned} $$

$\square$

Now back to the problem. The points from the above Lemma correspond in name with the points in the constellation

and we have shown the relation: $$ MX\cdot MP=MY\cdot MN\ . $$ From this, $PXYN$ is inscriptible. ($\Delta MXY\sim\Delta MNP$, $XY$ antiparallel to $NP$, $\angle MXY=\angle MNP$, $\angle MYX=\angle MPN$.)

We get now $$ \frac 12\operatorname{measure}(\overset\frown{MY}\text{ in circle }(O)) =\frac 12\widehat{YXM} =\frac 12\widehat{PNM} =\frac 12\widehat{NMC} =\frac 12\widehat{YMC} \ , $$ so $MC$ is tangent to $(O)$ in $M$.

$\square$

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Notes:

• The point $X$ constructed above is the center of the circumscribed circle of the triangle $\Delta O'BE$, since it lies on two perpendicular bisectors of the sides of the triangle,

  • the one side is $O'B$,
  • and the other side is $BE$, its perpendicular bisector being $MP$, since it is parallel to $AC\perp BE$, and goes (as mid line in the triangle) through the mid point of $BE$.

• The directions $XY\| FE$ and $PN\| BC$ are antiparallel, if considered with respect to the angle in $M$ formed by the half lines $MP$ and $MN$.

• An inversion with respect to $M$ may be the solution, since the Lemma isolates the power of the point $M$ with respect to the circle through $N,P,X,Y$. So the "complicated points" $X,Y$ become through this inversion the "simpler points" $NP$. For this inversion $W\to W'$, there correspond: $$ \begin{aligned} X&\to X'=P\ ,\\ Y&\to Y'=N\ ,\\ P&\to P'=X\ ,\\ N&\to N'=Y\ ,\\ \text{Circle }(MXY)&\to \text{Projective line }X'Y'=PN\ , \end{aligned} $$ and so on.