Let PH intersect BC at point X Let make Simmetry of line PH wrt sidelines of triangle ABC well known that they are intersects at point K' and K' is on (ABC) , so angle HXC = CXK Let P' is Simmetry of point P wrt BA Easy to see that K is on TK were T is intersection of (ABC) and CH Easy to see that HTPP' is Cyclic and so after Reim's Theorem MK' || HP' Well known that HP' || to Simson line of point P wrt ABC , so K' = K Let TK and PH intersected at point F on BA Let Circle (MXK) intersects line BC at points X and Y Easy to see that angle KYX = KMX = KTP Let line KY intersect (ABC) at point Q' , easy to see that angle KQ'P = KTP = KYB , so PQ' || BC , so Q' = Q KY = YM because MXKY is Cyclic and XY is angle bissector of MXK . done

Restate the problem as follow. Problem. Let $H$ be the orthocenter of an accute trangle $ABC$ with circumcircle $\Gamma$. Let $P$ be a point on the arc $AB$ (not containing $C$) of $\Gamma$, and let $M$ be a point on the arc $CA$ (not containing $B$) of $\Gamma$ such that $H$ lies on the segment $PM$. Let $K$ be another point on $\Gamma$ such that $KM$ is parallel to the Simson line of $P$ with respect to triangle $ABC$. Let $Q$ be another point on $\Gamma$ such that $PQ \parallel AB$. Segments $AB$ and $KQ$ intersect at a point $J$. Prove that $\triangle KJM$ is an isosceles triangle. Let $P_a$ , $P_b$, $P_c$ and $P'$ be reflections of $P$ across $BC$ , $AC$ , $AB$ and $UV$ respectively, where $U$ and $V$ are orthogonal projections of $P$ into $AC$ and $BC$. Then all those points belongs to line $P'H$ that is the steiner line of $P$ with respect to both of triangles $\triangle{ABC}$ and $\triangle{CUV}$. And since $(UV)\parallel (P'P_b)$, we get that $UV$ bissect $PH$ and $(UV)\parallel(HP_b)$. Let $B_1$ be the reflexion of $H$ across $AC$, $P_bB_1$ meet the circumcirlce again at $K'$, then $\angle{K'MH}=180-\angle{P_bB_1P}=\angle{P_bHM}$ so $(MK')\parallel(P_bH)\parallel(UV)$, thus $K'=K$. Thus $B_1K$ is the reflection of $PH$ across $AC$. So the reflection of $K$ with respect to $AC$ lies on $PH$ that is $PH$ is the steiner line of $K$ with respect to $\triangle{ABC}$, hence if $C_1$ is the refelection of $H$ across $AB$, lines $KC_1$ , $PH$ and $AB$ will be concourant. Now let $MJ$ meet the circumcircle of $\triangle{ABC}$ again at $R$. Using the lemma in Johan's post, we have : $QR.BK.AM=AQ.BR.MK$ and $MA.PC_1.BK=MK.AC_1.PB$ which leads to $\frac{RQ}{RB}=\frac{PC_1}{AC_1}$ and thus $ABRC_1$ is an isocel trapezoid. Let $HC_1$ me $UV$ and $AB$ at $C_2$ and $S$, $N$ the midpoint of $PH$, $W$ the projection of $P$ into $AB$. See that $HSPW$ is a parallelogram, so $\angle{SVC_2}=\angle{NC_2V}=\angle{PH_1}=\angle{QRH_1}$ and since $(RH_1)\parallel(AB)$ we conclude that $(RQ)\parallel(MK)$ impliying that $MQRK$ is an isocel trapezoid, Done!

let $PH$ intersect $BC$ at $R$, $AH$ intersect $\Gamma$ at $D$, and let $S$ be the intersection point of the perpendicular from $P$ to $BC$ with $\Gamma$. it's well known that $AS$ is parallel to the simson line of $P$ wrt $ABC$, so we have $AS||KM$. now let $DR$ intersect $\Gamma$ at $K'$. we have $RH=RD$, so $\angle RDA=\angle RPS$. so we conclude that $SM=AK'$, meaning that $MK'||AS$, so $K=K'$. now we have that $BC$ is the angle bisector of $\angle KRP$. $KMQP$ is cyclic, and $RJ||PQ$, so $KMJR$ is cyclic. now using this that $BC$ is the angle bisector of $\angle KRP$, we get that $JM=JK$, hence triangle $KJM$ is isosceles, and the proof is complete . (my proof is very short, am I missing something? )

No one posted a bash; hmm, let's change that. We will use one synthetic observation before laying down the hammer of complex bash. The Simson line is well known to be perpendicular to $AQ$, since it is perpendicular to the line joining $A$ and $P$'s isogonal conjugate, which is the point at infinity on $AQ$. Now we're ready for the bash. Let the perpendicular bisector of $MK$ be the real axis, so $m = \frac{1}{k}$. I claim that the condition $P, H, M$ collinear is equivalent to the concurrence of the perpendicular bisector of $MK$ with $BC$ and $AQ$. Clearly, this will imply the result. Observe that the intersection of the real axis with a chord $BC$ on the unit circle is $\frac{b+c}{bc+1}$; this is easy to show. Hence, the second condition is equivalent to \[ \frac{b+c}{bc+1} = \frac{k + q}{kq + 1}, \]which is equivalent to \[ bkq + ckq + b + c - kbc - k - qbc - q = 0 \]By the synthetic observation, one can check that $a = -\frac{1}{q}$. Now, $p = \frac{bc}{q}$ and $h = a + b + c = b + c - \frac{1}{q}$ and $m = \frac{1}{k}$, so $P$, $H$, $M$ collinear is equivalent to \begin{align*} \frac{p - h}{\overline{p} - \overline{h}} &= \frac{h - m}{\overline{h} - \overline{m}} \\ \frac{b + c - \frac{1}{q} - \frac{1}{k}}{\frac{1}{b} + \frac{1}{c} - q - k} &= \frac{\frac{bc}{q} - \frac{1}{k}}{\frac{q}{bc} - k} \end{align*} After expansion and cancellation, the above expression becomes \[ (bkq + ckq + b + c - kbc - k - qbc - q)(bck - q) = 0 \] We just need to show that $bck \neq q$ to finish it off, but observe that this implies $pk = 1$, but this is obviously false; as then $p = m$, which cannot be. Hence, we are done.

Once we guess the vertex of the isosceles triangle this is super-easy. (I cheated with geogebra for this part though ) Amir Hossein wrote: Let $H$ be the orthocenter of an acute trangle $ABC$ with circumcircle $\Gamma$. Let $P$ be a point on the arc $BC$ (not containing $A$) of $\Gamma$, and let $M$ be a point on the arc $CA$ (not containing $B$) of $\Gamma$ such that $H$ lies on the segment $PM$. Let $K$ be another point on $\Gamma$ such that $KM$ is parallel to the Simson line of $P$ with respect to triangle $ABC$. Let $Q$ be another point on $\Gamma$ such that $PQ \parallel BC$. Segments $BC$ and $KQ$ intersect at a point $J$. Prove that $\triangle KJM$ is an isosceles triangle. Let $P', H'$ be reflections of $P,H$ in $\overline{BC}$. Let $T=\overline{PH} \cap \overline{BC}$. Note that $\measuredangle KJT=\measuredangle KQP=\measuredangle KMP$ hence $KMJT$ is cyclic. It suffices to prove that $\overline{BC}$ bisects angle $PTK$. Observe that $\measuredangle PMK=\measuredangle PHP'=\measuredangle PH'P'$ hence $K$ lies on line $\overline{H'P'}$ as desired. $\blacksquare$ Note. It is tempting to try complex numbers and was the only trap in the problem for me

Let the reflection of $H$ across $BC$ be $G$. Let the perpendicular from $P$ onto $BC$ be $E$, and its extension onto the circumcircle be $U$. Finally, let $PM$ intersect $BC$ be $D$. Now, we shall try to phantom point to show that $K$ lies on $DG$ Let $K'$ be the intersection of $DG$ with the circumcircle. Firstly, observe that $HD=DG$ and $HG//UP$, which gives us the angle conditions $$\angle AGK'=\angle MPU$$ Hence, we get that the arc length $UM=AK'$. So, we have that $UA//K'M$ However, we know that by the Simson line config, $UA// simson line$ which implies that $K'M//Simson line$. However, it is given in the question that $KM//simsom line$, therefore we get the conclusion that $K=K'$ Since $$\angle DHC= \angle CDG= \angle BDP= \angle KDB$$, so BC bisects $\angle KDP$ Finally, note that $K,M,P,Q$ is cyclic and since $JD//PQ$, by Reim's, $K,M,D,J$ cyclic. Which implies that $$\angle JDP= \angle JKM= \angle JDK= \angle JMK$$Hence, $JM=JK$