Nice problem. An observation is that the pedal triangle of E wrt ABC is right angled.

Lemma (well-known) : Let $ P, $ $ Q $ be the isogonal conjugate WRT $ \triangle ABC. $ Then the insimilicenter, exsimilicenter of $ \odot (BPC) $ $ \sim $ $ \odot (BQC) $ lie on the circumcircle of $ \triangle ABC. $ Back to the main problem : Let $ G $ be the intersection of the segment $ AE $ and the circle $ \odot (T) $ with diameter $ BC. $ Since $ \measuredangle BEC $ $ + $ $ \measuredangle BGC $ $ = $ $ \measuredangle BAC, $ so from Lemma $ \Longrightarrow $ $ M, $ $ N $ is the insimilicenter, exsimilicenter of $ \odot (P) $ $ \sim $ $ \odot (T), $ respectively, hence we get $ EP $ is parallel to $ FT $ and $ ME $ $ \cdot $ $ MG $ $ = $ $ MN $ $ \cdot $ $ MT $ $ \Longrightarrow $ $ E, $ $ G, $ $ N, $ $ T $ are concyclic. Notice $ {PE}^2 $ $ = $ $ PM $ $ \cdot $ $ PN $ we get $$ \measuredangle (GT,AE) = \measuredangle TGE = \measuredangle TNE = \measuredangle MEP = \measuredangle (AE,FT) \Longrightarrow FG \perp AM, $$so the circumcircle $ \omega $ of $ \triangle FGM $ passes through the second intersection $ S $ of $ EF $ and $ \odot (ABC). $ Since the orthocenter $ H $ of $ \triangle EFM $ lies on $ FG, $ $ MS, $ so $ H $ is the radical center of $ \omega, $ $ \odot (T), $ $ \odot (ABC) $ $ \Longrightarrow $ $ H $ $ \in $ $ BC. $

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Thank you for your solution. This is the same as mine. For the ex- and in-similicenter of two circles $(P) $ and $(BC)$, just note that $\angle MCB=\angle MAB=\angle MCP$ hence $CM$ is the bisector of $\angle BCP.$ Therefore $CN$ is the external bisector of $\angle BCP.$

Generation Let $ABC$ be a triangle inscribed in $(O)$. Two tangents of $(O)$ at $B,C$ meets at $P$. Let $M,N$ be the midpoints of arcs $BC$ and $BAC$. $K$ is a point on the perpendicular bisector of $BC$ such that $AK$ cuts $(P)$. $(P,PA)$ cuts $AK$ at $E,$ $(BC)$ cuts $EN$ at $F.$ Prove that the orthocenter of triangle $EFM$ lies on $BC$. Proof Let $J$ be the midpoint of $BC, ET$ be the diameter of $(P)$, $PM$ cut $(P)$ at $R$, cut $(EN)$ at $F$. Then we will prove that $F$ lies on $(J,JB).$ Let$ EN $cut $(O)$ at $I,$ $MI$ cut $BC$ at $S$, then $\angle NIM=90^o$ and $S$ lies on $ER$. Invertion with center $M$, radius $MB^2$, then $MI.MS=MR.MF=MI.MN$ $ \Longrightarrow $ $FRJN$ is concyclic, so $\angle RFJ=\angle RNJ$. But we well-known that $MP.MN=MR.MT$, then $NRPT$ is concyclic $ \Longrightarrow $ $\angle RNJ=\angle MTP$, so $FJ\parallel EP$ $ \Longrightarrow $ $\frac {FJ}{ET}=\frac {NJ}{NP}$$ \Longrightarrow $ $FJ=$$\frac {1}{2}.BC$, so $F$ lies on $(BC)$. And note that $S$ is the othocenter of $\triangle EFM$, then $S$ lies on $BC.$

doxuanlong15052000 wrote: Generation Let $ABC$ be a triangle inscribed in $(O)$. Two tangents of $(O)$ at $B,C$ meets at $P$. Let $M,N$ be the midpoints of arcs $BC$ and $BAC$. $K$ is a point on the perpendicular bisector of $BC$ such that $AK$ cuts $(P)$. $(P,PA)$ cuts $AK$ at $E,$ $(BC)$ cuts $EN$ at $F.$ Prove that the orthocenter of triangle $EFM$ lies on $BC$. Let $ T $ be the center of $ \odot(BC). $ From Lemma at post #3 $ \Longrightarrow $ $ M $ is the insimilicenter of $ \odot (P) $ $ \sim $ $ \odot (T), $ so there is a homothety $ \mathbb{H}_M $ with center $ M $ that maps $ \odot (T) $ to $ \odot (P). $ Similarly, $ N $ is the exsimilicenter of $ \odot (P) $ $ \sim $ $ \odot (T) $ and there is a homothety with center $ N $ that maps $ \odot (P) $ to $ \odot (T). $ Since $$ \mathbb{H}_M(\mathbb{H}_N(E)) = \mathbb{H}_M(F) = \text{antipode of E in} \odot (P), $$so the projection $ S $ of $ E $ on $ FM $ lies on $ \odot (P). $ Clearly, the circumcircle $ \omega $ of $ \triangle EMS $ passes through the 2nd intersection $ J $ of $ EF $ and $ \odot (O), $ so the orthocenter $ H $ $ \equiv $ $ ES $ $ \cap $ $ JM $ of $ \triangle EFM $ (radical center of $ \omega, $ $ \odot (O), $ $ \odot (P) $) lies on $ BC. $

One more observation. Let $D$ be the second intersection of $AP$ and $(O)$, $L$ be the intersection of $NE$ and $(O)$. Then $(LED)$ is tangent to $(P,PB).$

livetolove212 wrote: Let $D$ be the second intersection of $AP$ and $(O)$, $L$ be the intersection of $NE$ and $(O)$. Then $(LED)$ is tangent to $(P,PB).$ Let $ O, $ $ T $ be the midpoint of $ MN, $ $ BC, $ respectively. Clearly, $ ET, $ $ EO $ is the E-symmedian of $ \triangle EMN, $ $ \triangle BCE, $ respectively, so $ EM, $ $ EN $ are isogonal conjugate WRT $ \angle BEC. $ Let $ \tau_E $ be the tangent of $ \odot (BCE) $ through $ E. $ From $ \measuredangle PDN $ $ = $ $ \measuredangle EMN $ $ = $ $ \measuredangle PEN $ $ \Longrightarrow $ $ D, $ $ E, $ $ N, $ $ P $ are concyclic, so we get $$ \measuredangle EDL = \measuredangle EDP + \measuredangle PDL = \measuredangle LNM + \measuredangle ANL = \measuredangle ANM = \measuredangle (AE,BC) = \measuredangle (\tau_E ,EL) $$$ \Longrightarrow $ $ \tau_E $ is tangent to $ \odot (DEL) $ at $ E, $ hence we conclude that $ \odot (P) $ and $ \odot (DEL) $ are tangent to each other at $ E. $

livetolove212 wrote: Let $ABC$ be a triangle inscribed in $(O)$. Two tangents of $(O)$ at $B,C$ meets at $P$. The bisector of angle $BAC $ intersects $(P,PB)$ at point $E$ lying inside triangle $ABC$. Let $M,N$ be the midpoints of arcs $BC$ and $BAC$. Circle with diameter $BC$ intersects line segment $EN$ at $F$. Prove that the orthocenter of triangle $EFM$ lies on $BC$. $AM$ cuts $(BC)$ at $D$ and cuts $(P,PB)$ at $G$. Taking inversion centre $A$ , abitrary power and reflect about $AM$ so $B \rightarrow B, C \rightarrow C , (O) \leftrightarrow (P) $ so $D \leftrightarrow G$. Then $D,E$ are isogonal conjugate in $\triangle ABC$ $\implies MB^2 = MD.ME = MX.MN$ ($X$ is the midpoint of $BC$) then $NDEX$ is cyclic. Note that $PE^2 = PM.PN$ and $(NMXP) = -1 $ so $\frac{NX}{NP} = \frac{MN}{MP} = \frac{BX}{BP}$ . Then we have $PE \parallel XF$ . $XF$ cuts $AD$ at $Z$ , then we have : $\angle EDX = \angle ENX = \angle PEM = \angle EZX$ so $XD = XZ$. Then $Z$ lies on $(BC)$ or $\angle ADF = 90$ $NF$ cuts $(O)$ again at $Y$ then $\angle NYM = 90$ . By applying the corollary of radical-axis for $(O)$, $(BC)$ and $(FDYM)$, we get the orthocenter of $\triangle EFM$ lies on $BC$

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livetolove212 wrote: One more observation. Let $D$ be the second intersection of $AP$ and $(O)$, $L$ be the intersection of $NE$ and $(O)$. Then $(LED)$ is tangent to $(P,PB).$ $PE^2 = PD.PA$ so $PD$ is the tangent of $(AED)$. Then $\angle DEP = \angle DAE $. Let $d$ be the tangent at $E$ of $EDL$. Then we have : $\angle dEP = \angle dED - \angle DEP = \angle ELD - \angle DEP = \angle NAD - \angle DEP = \angle NAE = 90$ . Then $d$ is also the tangent of $P$ . Then we have $(P)$ and $(EDL)$ tangent at $E$ An other extension : $\triangle ABC$, $P$ is the intersection of 2 tangents at $B,C$ of $(O)$. $M,N$ are the midpoints of arc $BC$ and $BAC$. $E$ is an abitrary point of $(P;PB)$ , $NE$ cuts $(BC)$ at $F$ then orthocenter of $\triangle EFM$ lies on $BC$