Post Number:**#2** by **Fm Jakaria** » Sat Aug 01, 2015 8:31 pm

Let $O$ be the center of $\gamma$, the circumcircle of ABC. $ AH = 2OM$, because$ O$ is the orthocenter of the medial triangle of $\triangle ABC$. So $P := AO \cap HM$ sends $OM$ to $AH$ by a homothety of ratio $2:1$. So $P$ lies on $\gamma$, and it is diametrically opposite to $ A$. As the projection of $A$ on $HP$ lies on $\gamma$; $Q$ coincides with this projection, so Q lies on $PMH$.

Let $AHF$ intersect $\gamma$ again at $R$. As $\angle HCF = \angle FCR = 90-B$, $HR$ has perpendicular bisector $BC$.

Let segment $KR$ intersect the circumcircle $\omega$ of $QKH$ again at $X$.

Note $ \angle HXK = 180 - \angle PQK = \angle PRK$. So $HX$ and $PR$ are parallel, so both is perpendicular to $AHFR$, and so both is parallel to $BC$. $HQ$ is a diameter of $\omega$, so $QX$ is perpendicular to $HX$, so to $BC$. Let us extend $QX$ to meet $PR$ at $Y$, and $BC$ at $L$. Now it is obvious that $HRYX$ is a rectangle, with $BLC$ the perpendicular bisector of $XY$.

Let the perpendicular to $QHP$ at $H$ meet $BC$ at $Z$. Let $OZ$ meet $RXK$ at $S$.

Now note that because $HX$ is parallel to $MZ$, the right triangles $MHZ$ and $HXQ$ are similar. So

$\dfrac {HX}{MH} = \dfrac{HQ}{MZ}$. Moreover, $AQFM$ is cyclic with diameter $AM$. So

$AH.HF = MH.HQ$. Combining the last two gives

$HX.(MZ) = \dfrac{1}{2}.AH.(2HF) = OM. HR$. So $ \dfrac {HX}{HR} = \dfrac{OM}{MZ}$, so

$OMZ \sim XHR \sim RYX$. This gives $\angle LZS = \angle MZO = \angle RXL$. So $XLZS$ is cyclic. Then $\angle XSZ = \angle XLZ = 90$, so $OZ$ is perpendicular to $RXK$. Since $OR = OK$, we have that $ZK = ZR = ZH$.

Note that the circumcircle of $HMF$ and $\omega$ has the common tangent $ ZH$ at $H$; because they have the respective diameters HM and HQ. This gives, as $ZH = ZK$, that $ZK$ is tangent to $\omega$. Moreover, $ ZK^2 = ZH^2 = ZF.ZM$, giving that $ZK$ is also tangent to the circumcircle of $KMF$. So give a big clap….