khashi70 wrote: In triangle $ ABC$ , $ D$ , $ E$ and $ F$ are the points of tangency of incircle with the center of $ I$ to $ BC$ , $ CA$ and $ AB$ respectively . Let $ M$ be the feet of perpendicular from $ D$ to $ EF$ and $ P$ is on $ DM$ such that : $ DP = MP$ . If $ H$ be the orthocenter of $ ABC$, prove that $ PH$ bisects $ EF$. Maybe you have made a typo, check it again.

Sorry you're right ... it's OK now .

Let $ CI$ cut $ BH$ at $ G$ and $ BI$ cut $ EF$ ate $ Q$. Then $ \angle GIB = 180^\circ - \dfrac{\angle B + \angle C}{2} = 90^\circ - \dfrac{A}{2} = \angle EFA$. Thus, the quadrilateral $ BFGI$ is cyclic and $ \angle{BGI} = \angle BFI = 90^\circ$. So, points $ B,G$ and $ H$ are collinear. Denote $ K \in HI \cap EF$. Let $ J$ be midpoint of $ EF$. By Menelaos theorem for $ \triangle DMK$ and points $ P,J,H$ it's enough to prove that $ \dfrac{PM}{PN}\cdot \dfrac{HD}{HK}\cdot \dfrac{JK}{JM} = 1 \Longleftrightarrow \dfrac{HD}{HK} = \dfrac{JM}{JK} \ (*)$. From $ AF = AE$ we get $ IJ \perp EF$, so $ JI \parallel DM$. Therefore $ \dfrac{JM}{JK} = \dfrac{ID}{IK}$. We rewrite the relation $ (*)$ as $ \dfrac{HD}{HK} = \dfrac{ID}{IK}$ This is equivalent to prove that $ (D,I,K,H)$ is harmonical. $ QGD$ is orthic triangle of $ HBC$. So $ GI$ is bisector of $ DGK$. From $ \angle IGH = 90^\circ$ we get that $ GH$ is exterior bisector of $ DGK$ and the result follows. Image not found

Sketch of my solution 1. The intersection points of the altitudes from B and C with the corresponding sides in triangle HBC are on the line EF (well known lemma) 2. H, I and D are collinear. (easy with the use of harmonical division in both triangles ABC and HBC) 3. HBC and DEF are similar. (angle chasing) 4. Let the orthocenter of DEF be H' and the midpoint of EF be N. Now from the well known property, H'D is equal to 2 times IN. Using 3, we find that AI/AD=IN/PD. Then the result follows from the fact that IN//PD.

Here is another aproach for this problem: Let $ \{K\}\equiv CI\cap FE$, $ \{G\}\equiv BI\cap EF$. It is very well- know that $ BK\bot CK$ and $ BG\bot CG$ (this result can be proved easily by harmonic division). Hence $ \{H\}\equiv BK\cap CG$. Let $ J$ be the midpoint of $ EF$. Denote $ P$ the intersection by $ HJ$ and $ DM$. We will prove that $ P$ is the midpoint of $ DM$. Indeed, let $ S$ be the projection of $ H$ onto $ EF$, in order to prove $ P$ is the midpoint of $ DM$, it is enough to prove $ H(MJYS) = - 1$, where $ \{Y\}\equiv HD\cap EF$, which is equivalent to proving $ \frac {\overline {JM}}{\overline {JY}} = - \frac {\overline {SM}}{\overline {SY}}$. Note that $ J$ is the midpoint of $ EF\Longrightarrow IJ\bot EF\Longrightarrow IJ//DM$, which implies $ \frac {\overline {JY}}{\overline {JM}} = \frac {\overline {IY}}{\overline {ID}}$. But $ (HIYD) = - 1\Longrightarrow \frac {\overline {IY}}{\overline {ID}} = - \frac {\overline {HY}}{\overline {HD}} = - \frac {\overline {HY}}{\overline {HY} + \overline {YD}} = - \frac {\overline {SY}}{\overline {SY} + \overline {YM}} = - \frac {\overline {SY}}{\overline {SM}}$ $ \Longrightarrow \frac {\overline {SY}}{\overline {SM}} = - \frac {\overline {JY}}{\overline {JM}}$. Which implies that $ (MJYS) = - 1\Longrightarrow H(MDPS) = - 1$, but $ DM//HS\Longrightarrow P$ is the midpoint of $ DM$, which leads to the result of the problem. Our proof is completed then.

Sorry for my bad English... Here is the proof without calculation. Lemma:In a triangle ABC,let D be the tangent of incircle $ (I)$ to BC,$ (I_a)$ be the excircle of A wrt triangle ABC and M be midpoint of the altitude from A wrt triangle ABC,then $ D,I_a,M$ are collinear. Proof: Let $ (I_a)$ meet BC at N,$ NI_a$ intersect $ (I_a)$ at N and P The homothety center A take $ (I)$ to $ (I_a)$ also take D to P,then A,D,P are collinear. Then the homothety center D take the altitude from A wrt triangle ABC to NP,it also take M to $ I_a$ We conclude that $ D,I_a,M$ are collinear. Return to the problem,apply the lemma to the orthic triangle of triangle BHC then P,H,M' are collinear (M' is the midpoint of EF). Q.E.D

let $ K$ midpoint of $ EF$ and $ r$ radius of incircle of $ \triangle ABC$. let $ PH$ cut $ IK$ at $ K'$ so $ \frac{IK'}{DP}=\frac{HI}{HD}$ ($ i$)(because of $ IK || PD$) it is easy to measure this lenght: $ IK=r \sin\frac{A}{2}$ (1) $ DM=BD\cos\frac{B}{2}=2r\cos\frac{c}{2}\cos\frac{B}{2}$ so $ DP=r\cos\frac{B}{2}\cos\frac{C}{2}$ (2) with (1),(2): $ \frac{IK}{DP}=\frac{\sin\frac{A}{2}}{\cos\frac{B}{2}\cos\frac{C}{2}}$ (3). and $ ID=BD \tan\frac{B}{2}$,$ DH= \cot\frac{C}{2}$ so $ HI=BD(\cot\frac{C}{2}-\tan\frac{B}{2})$ so $ \frac{HI}{DH}=1-\frac{\tan\frac{B}{2}}{\cot\frac{C}{2}}=1-\tan\frac{B}{2}\frac{C}{2}=\frac{\sin\frac{A}{2}}{\cos\frac{B}{2}\cos\frac{C}{2}}$ (4). with (3),(4): $ \frac{IK}{DP}=\frac{HI}{HD}$ ($ ii$). with ($ i$),($ ii$): $ PH$ passer from $ K$. problem solved. Image not found

let $T$ be the intersection point of $PH$ and $EF$, $S$ the intersection point of $IH$ and $EF$, and $R,T'$ be the feet of the perpendicular from $H$ and $I$ to the line $EF$ respectively. since $HR||DM$ and $P$ is the midpoint of $DM$, we have $H(RPDM)=-1$, so $(RTSM)=-1$. on the other side, $EF$ is the line joining feet of the altitudes of vertices $B$ and $C$ of triangle $HBC$ respectively, so $(HISD)=-1$ and therefore $(RT'SM)=-1$. we cocnlude that $T=T'$, hence $T$ is the midpoint of $EF$, and the proof is complete.

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(Writing the Idea only, calculation is easy).Take $F=(1,0),E=(-1,0),I=(0,r),M=(a,0),D=(a,b)$ now so equation of circle $\odot(DEF)$ be $x^2+(y-r)^2=r^2+1$. Now note tangent at $(x_1,y_1)$ point be $xx_1+yy_1-r(y+y_1)=1$, using this we get $B,C$ in terms of $a,b,r$. Now easily we’ll get $H$ in terms of $a,b,r$. We’ve midpoint of $MD$ is $P=(a,\frac {b}{2})$ , now basically we need to show now slope of the line is $tan^{-1}(\frac {b}{2})$ and that’s quite easy to show. Note here we’re saying , solution is easy by CG because no stuffs involving any power is coming, and that makes our calculation easier. (I got a trig solution too, but there calculation is damn hard to do.)

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.535709806738513, xmax = 7.699401969463958, ymin = 1.537544229253762, ymax = 7.025735038013770; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((4.120000000000004,6.740000000000007)--(0.6000000000000006,2.120000000000002)--(4.046266301600477,2.770413386903411)--cycle, zzttqq); /* draw figures */ draw((4.120000000000004,6.740000000000007)--(0.6000000000000006,2.120000000000002), zzttqq); draw((0.6000000000000006,2.120000000000002)--(4.046266301600477,2.770413386903411), zzttqq); draw((4.046266301600477,2.770413386903411)--(4.120000000000004,6.740000000000007), zzttqq); draw((2.219650741964543,4.245791598828465)--(4.061766128857575,3.604874537240598)); draw((3.226139448965370,2.615630953617537)--(3.623387848646997,3.757397078019321)); draw((0.6000000000000006,2.120000000000002)--(2.823005108246210,4.751669850437923)); draw((3.424763648806184,3.186514015818429)--(2.823005108246210,4.751669850437923)); draw((3.623387848646997,3.757397078019321)--(4.046266301600477,2.770413386903411)); draw((2.823005108246210,4.751669850437923)--(4.046266301600477,2.770413386903411)); draw((3.623387848646997,3.757397078019321)--(4.046266301600477,2.770413386903411)); draw((4.046266301600477,2.770413386903411)--(2.355699169206497,4.198457018789681)); draw((0.6000000000000006,2.120000000000002)--(3.385861897857764,3.840038173964708)); draw((3.035843169874220,3.623930701373472)--(2.823005108246210,4.751669850437923)); draw((3.226139448965370,2.615630953617537)--(3.035843169874220,3.623930701373472)); /* dots and labels */ dot((4.120000000000004,6.740000000000007),dotstyle); label("$A$", (4.153046114633986,6.797820137317756), NE * labelscalefactor); dot((0.6000000000000006,2.120000000000002),dotstyle); label("$B$", (0.6340400478875363,2.175705951202600), NE * labelscalefactor); dot((4.046266301600477,2.770413386903411),dotstyle); label("$C$", (4.080113346411262,2.822984269179279), NE * labelscalefactor); dot((3.035843169874220,3.623930701373472),dotstyle); label("$I$", (3.068171187320962,3.679944295796290), NE * labelscalefactor); dot((2.219650741964543,4.245791598828465),dotstyle); label("$F$", (2.256794140843153,4.299872825689447), NE * labelscalefactor); dot((3.226139448965370,2.615630953617537),dotstyle); label("$D$", (3.259619703905613,2.668002136705990), NE * labelscalefactor); dot((4.061766128857575,3.604874537240598),dotstyle); label("$E$", (4.098346538466943,3.661711103740609), NE * labelscalefactor); dot((3.623387848646997,3.757397078019321),dotstyle); label("$M$", (3.660749929130597,3.807576640186058), NE * labelscalefactor); dot((3.424763648806184,3.186514015818429),dotstyle); label("$P$", (3.460184816518106,3.242347686459944), NE * labelscalefactor); dot((2.823005108246210,4.751669850437923),dotstyle); label("$H$", (2.858489478680629,4.810402203248517), NE * labelscalefactor); dot((3.140708435411058,3.925333068034533),dotstyle); label("$N$", (3.177570339655048,3.980791964715028), NE * labelscalefactor); dot((2.355699169206497,4.198457018789681),dotstyle); label("$J_c$", (2.393543081260761,4.254289845550244), NE * labelscalefactor); dot((3.385861897857764,3.840038173964708),dotstyle); label("$J_b$", (3.423718432406743,3.898742600464463), NE * labelscalefactor); dot((2.967592072231618,3.985564490881980),dotstyle); label("$X$", (3.004355015126078,4.044608136909912), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $J_b$ and $J_c$ be the feet of perpendiculars from $B,C$ on $CH,BH$ respectively. Since $CI \perp BH$ and $BI \perp CH,$ $B,I,J_b$ and $C,I,J_c$ are collinear. Furthermore, since $\angle BJ_cC = \angle BJ_bB = 90^{\circ},$ $(BI_cI_bC)$ is cyclic and since $ID \perp BD, IF \perp BF, IJ_c \perp BJ_c,$ $(BDIJ_cF)$ is also cyclic. Now, note that $\angle FJ_cB = \angle FDB = 90^{\circ} - \dfrac{\angle ABC}{2}$ and $\angle J_bJ_cC = \angle J_bBC = \dfrac{\angle ABC}{2},$ so $\angle \angle FJ_cB + \angle BJ_cC + \angle J_bJ_cC = 180^{\circ},$ implying $F,J_c, J_b$ are collinear. Similarly, $E, J_b, J_c$ are collinear. Combining them, we see that $J_b$ and $J_c$ both lie on line $EF.$ Since $ID \perp BC,$ $H,I,D$ are collinear. Let this line intersect $EF$ at $X.$ Using cyclic quadrilaterals $BDJ_bH$ and $CDJ_cH,$ we deduce that \[\angle J_cJ_bI = 90^{\circ} - \angle HJ_bJ_c = 90^{\circ} - \angle CJ_bE = \angle IJ_bC,\] so $J_bI$ internally bisects $\angle J_cJ_bC.$ Since $IJ_b \perp J_bH,$ $J_bH$ externally bisects $\angle J_bJ_bC,$ so by the angle bisector theorem we have $\dfrac{HD}{ID} = \dfrac{HX}{IX}.$ Let $N$ be the midpoint of $EF.$ Note that $A$ and $I$ both lie on the perpendicular bisector of $EF,$ so $IN \perp EF,$ implying $(H,I;X,D)$ is harmonic, or $\dfrac{KX}{IX} = \dfrac{DH}{DI}.$ Plugging these ratios, a simple calculation shows that \[\dfrac{PM}{PN} \cdot \dfrac{DH}{DX} \cdot \dfrac{XN}{MN} = 1,\] so the desired result follows by the converse of Menelaus theorem. $\blacksquare$

I will provide a more straightforward and bashy solution.(but a synthetic solution like you have done is obviously more appreciated) Applying the sine rule in $\triangle{BFD}$ we get $\frac{FD}{sinB}=\frac{s-b}{cos\frac{B}{2}} \implies FD=2sin\frac{B}{2}(s-b)$ So $MD=FDcos\frac{C}{2}=2sin\frac{B}{2}cos\frac{C}{2}(s-b) \implies DP=sin\frac{B}{2}cos\frac{C}{2}(s-b)$ Let $K$ be the midpoint of $EF$.Then $IK=IEsin\frac{A}{2}=rsin\frac{A}{2}=(s-b)tan\frac{B}{2}sin\frac{A}{2}$ Now easy angle chasing gives $\angle{ICH}=\frac{A}{2}$ and $\angle{HIC}=90+\frac{C}{2}$ So applying sine rule in $\triangle{HIC}$ we get $HI=\frac{HCsin\frac{A}{2}}{cos\frac{C}{2}}$ Also $HD=HCcos\frac{B}{2}$. Now note that $\frac{IK}{PD}=\frac{(s-b)tan\frac{B}{2}sin\frac{A}{2}}{sin\frac{B}{2}cos\frac{C}{2}(s-b)}=\frac{sin\frac{A}{2}}{cos\frac{B}{2}cos\frac{C}{2}}$ Again $\frac{HI}{HD}=\frac{\frac{HCsin\frac{A}{2}}{cos\frac{C}{2}}}{HCcos\frac{B}{2}}=\frac{sin\frac{A}{2}}{cos\frac{B}{2}cos\frac{C}{2}}$ Thus $\frac{HI}{HD}=\frac{IK}{PD}$.Also note that $IK \parallel PD$,both being perpendicular to $EF$. So $H,K,P$ are collinear as desired.

My solution: Let $BI\cap EF=T$ note that $\angle BTF=\angle AFE-\angle ABI=\frac{C}{2},\angle ECI=\frac{C}{2}\longrightarrow IETC$ is cyclic so $\angle IEC=\angle ITC=90$ thus $C,T,H$ are collinear similarly $B,S,H$ are collinear now let the perpendicular lines from $D$ to $DE,DF$ intersect $EF$ at $K,L$ note that $BT||DL$ (because both of them are perpendicular to $HC$) and $\angle STB=\angle BTD=\frac{C}{2}$ thus $TL=TD$ so $TH$ is perpendicular bisector of $DL$ similarly $HS$ is perpendicular bisector of $DK$ so $H$ is circumcenter of $\triangle DKL$ now using this problem we get that $P,M,H$ are collinear. DONE

Let $Q=BI \cap EF$, $N=AI \cap EF$, and $K=BI \cap DF$. Since $\angle CIQ=\angle CBI + \angle ICB = 90^{\circ} - \angle IAE = \angle AEF = \angle CEQ$ and $90^{\circ}=\angle CDI = \angle IEC$ the pentagon $CDIEQ$ is cyclic with $CI$ as diameter, so $\angle IQC = 90^{\circ}$ and hence $Q \in CH$. Denote by $h$ the homothety centered at $H$ sending $I$ to $D$, and let $R=h(Q)$. We claim that $h(N)=P$ from which the assertion follows since $N$ is also the midpoint of $EF$. But obviously $h(IN)=DP$ since $IN \parallel DP$, so we just have to show that $h(QN)=RP$. Note first that $h(IQ)=DR$ so $IQ \parallel DR$ which means the pentagon $DKMQR$ is cyclic since $90^{\circ} = \angle QRD = \angle DKQ = \angle DMQ$. Note also that $KP$ is the perpendicular bisector of $DM$ since it is the midline of the right angled triangle $DFM$, so $ \angle DKR = 90^{\circ} - \angle KRD = 90^{\circ} - \angle KMD = \angle DKP$ which yields $R \in KP$, but $KP\parallel FM$ so we are done.

Suppose $X,Y$ are points on $BI,CI$ with $\angle BXC=\angle BYC=90$. It is well known that $X$ and $Y$ lie on $EF$. Let lines $DIH$ and $AI$ intersect $EF$ at $Z$, $W$; then $(H,I;D,Z)$ is harmonic. Projecting from $W$ yields $(WH,WI;WM,WD)$ is harmonic, but $WI\parallel DM$, so $HW$ bisects $DM$ as desired.

The given problem is equivalent to showing the following problem : Let $ABC$ be a triangle and let $AD$ be $A-$altitude ($D$ on $BC$) and let $P$ be the midpoint of $AD$. Let $M$ be the midpoint of $BC$. Let $O$ be the circumcentre of $\triangle ABC$ Let the $A-tangent$ to $\odot ABC$ be $l_A$; $l_B,l_C$ are defined similarly. $A'=l_B \cap l_C$ $B',C'$ are defined similarly. let $K$ be the Lemoine point of $\triangle ABC$ let $AO \cap PM = X$ Then $X$ is the orthocentre of triangle $B'OC'$ Solution : We know that $PKM,BKB',CKC'$ are straight lines. if we can show that $B'X||AB$ (and similarly $C'X||AC$), we will be done. Thus if we show $\triangle XAB'$ and $\triangle CMO$ are similar, we are done. That is we have to show, $\frac{B'A}{OM} = \frac{AX}{CM}$ By Menalu's theorem; $\frac{AX}{OX}\cdot \frac{OM}{MA'}\cdot \frac{A'K}{KA} = -1$ Since we know all these ratios, the required fractions can be easily shown equal

A simple complex bash...Take the in-circle to be the unit circle and observe that If $X$ is the midpoint of $EF$ then $X \longrightarrow x=\frac{e+f}{2}$ and the orthocenter of $\triangle BIC$ is $H\longrightarrow h=\frac{2d^2(e+f)}{(d+e)(d+f)}$ and mid point of altitude $DM$ is $P\longrightarrow p=3d/4+(e+f)/2-ef/d$ Now applying the criterion of col-linearity in complex numbers we conclude.

of $\triangle DEF.$ Let $\mathcal{H}$ be the circumconic of $\triangle ABC$ passing through $G, I$ and let this line meet $\mathcal{H}$ for a second time at $H'.$ Denote $X \equiv BH' \cap CI$ and $Y \equiv CH' \cap BI.$ It's sufficient to show that $H'$ is the orthocenter of $\triangle BIC.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 939.9590493942901, xmax = 1098.890727397895, ymin = 973.1890994475792, ymax = 1053.622348663321; /* image dimensions */ draw((978.0750118233286,1041.1012964658203)--(972.3534714152007,984.1622953027858)--(1046.4294535107942,984.438698221053)--cycle, red); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((978.0750118233286,1041.1012964658203)--(972.3534714152007,984.1622953027858), red); draw((972.3534714152007,984.1622953027858)--(1046.4294535107942,984.438698221053), red); draw((1046.4294535107942,984.438698221053)--(978.0750118233286,1041.1012964658203), red); draw(circle((993.5398909054971,1003.3984720933333), 19.156989626299726), dotted); draw((993.6113585293933,984.2416157770934)--(983.4573969568015,1008.9966124261042), green); pair hyperbolaLeft1 (real t) {return (18.787870262588065*(1+t^2)/(1-t^2),18.787870262588065*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (18.787870262588065*(-1-t^2)/(1-t^2),18.787870262588065*(-2)*t/(1-t^2));} draw(shift((992.0859932484519,1022.5002111017292))*rotate(-77.70722080273299)*graph(hyperbolaLeft1,-0.99,0.99), linetype("2 2") + blue); draw(shift((992.0859932484519,1022.5002111017292))*rotate(-77.70722080273299)*graph(hyperbolaRight1,-0.99,0.99), linetype("2 2") + blue); /* hyperbola construction */ draw((988.5343777430974,996.6191141015988)--(993.3926741433031,1042.8525643613414), green); draw((974.4788921062582,1005.3138248852948)--(1013.0946180629153,1021.1531563552602), green); /* dots and labels */ dot((978.0750118233286,1041.1012964658203),linewidth(3.pt) + dotstyle); label("$A$", (976.7206375238196,1042.4280304735012), N * labelscalefactor); dot((972.3534714152007,984.1622953027858),linewidth(3.pt) + dotstyle); label("$B$", (972.2981908315454,981.0665826181931), S * labelscalefactor); dot((1046.4294535107942,984.438698221053),linewidth(3.pt) + dotstyle); label("$C$", (1045.8213670906043,981.6193884547274), S * labelscalefactor); dot((993.6113585293933,984.2416157770934),linewidth(3.pt) + dotstyle); label("$D$", (993.0284097015808,981.4811869955938), S * labelscalefactor); dot((1005.7657262922834,1018.147004245295),linewidth(3.pt) + dotstyle); label("$E$", (1006.0193468601364,1019.4865882573275), NE * labelscalefactor); dot((974.4788921062582,1005.3138248852948),linewidth(3.pt) + dotstyle); label("$F$", (972.2981908315454,1005.5282408848362), NW * labelscalefactor); dot((993.5398909054971,1003.3984720933333),linewidth(3.pt) + dotstyle); label("$I$", (994.1340213746494,1004.284427752634), N * labelscalefactor); dot((989.0051739983018,1001.0992924544519),linewidth(3.pt) + dotstyle); label("$G$", (987.0857469588374,1002.4878087838975), N * labelscalefactor); dot((993.3926741433031,1042.8525643613414),linewidth(3.pt) + dotstyle); label("$H'$", (993.1666111607144,1044.2246494422377), N * labelscalefactor); dot((983.4573969568015,1008.9966124261042),linewidth(3.pt) + dotstyle); label("$M$", (983.2161061030974,1010.3652919545115), NE * labelscalefactor); dot((988.5343777430974,996.6191141015988),linewidth(3.pt) + dotstyle); label("$P$", (986.8093440405702,994.7485270724172), S * labelscalefactor); dot((980.8765540564431,1007.9380067633089),linewidth(3.pt) + dotstyle); label("$X$", (979.2082637882239,1008.9832773631756), NE * labelscalefactor); dot((1013.0946180629153,1021.1531563552602),linewidth(3.pt) + dotstyle); label("$Y$", (1013.6204271124827,1021.9742145217318), NE * labelscalefactor); dot((990.1223091992708,1011.730414565295),linewidth(3.pt) + dotstyle); label("$N$", (991.2317907328444,1013.129321137183), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] By Pascal's Theorem for $CGH'BAI$, it follows that $F, N, X$ are collinear. Therefore, $X$ is a common point of $EF$ and $CI$, so it's well-known (Lemma 8) that $\angle BXC = 90^{\circ}.$ Similarly, $\angle BYC = 90^{\circ}.$ Therefore, $H'$ is the orthocenter of $\triangle BIC$, as desired.

Considering the polar dual of what we need to show that $AI_B , I_AX $ and the $A-$midline of $ABC$ are concurrent where $X$ is such that $XI \perp IA$ and $X \in BC$ . Now , we bary it .Take $ABC$ as reference triangle . We only do not know $X$ which turns out as $X=(0,k,1-k) $ where $k=\frac{ab+bc+ca}{(a+c)(a+b+c)}$ . The remaining computations are trivial . Hence , done !

Let $Z$ be the midpoint of $EF$ and $P'=DM\cap ZH$. From the ceva-menelaus configuration we have \[ -1=(H, I; EF\cap DH, D)\stackrel{Z}{=}(P', \infty; M, D) \]Means $P'$ is the midpoint of $DM$. Hence $P'\equiv P$.

Let $X = \overline{BI} \cap \overline{EF}$ and $Y = \overline{CI} \cap \overline{EF}$, and denote the midpoint of $\overline{EF}$ as $N$. By Iran Lemma, $X$ and $Y$ are the feet of the $C$ and $B$ altitudes in $\triangle BIC$, respectively. So, $\triangle DXY$ is the orthic triangle of acute $\triangle BHC$, hence $I$ is its incenter and $H$ is its $D$-excenter. Then by the Midpoint of altitudes lemma, we find that $P$, $N$, and $H$ are collinear since $N$ is the foot from $I$ onto $\overline{XY}$, as desired.

Let $BI\cap CH = K$, $CI\cap BH = L$, and let $N$ be the midpoint of $EF$. Note that $I$ is the antipode of $H$ in $(HKL)$, and $K$ and $L$ are on $EF$ by Iran Lemma. Then since $\triangle DEF$ and $\triangle HLK$ are homothetic, and if $M'$ is the reflection of $M$ over $N$, $DM'$ is the same cevian in $\triangle DEF$ as $HN$ in $\triangle HKL$ (isotomic to altitude), we have $PN\parallel DM'\parallel HN$ as desired.

Let $T_B$ and $T_C$ denote the feet from $B$ to $CH$ and the feet from $C$ to $BH$ respectively. Note that $BICH$ is an orthocentric system, so $B,I,T_B$ are collinear as well as $C,I,T_C.$ By that egmo lemma, $BI$ and $EF$ intersect at some point $K$ for which $KB\perp CK.$ Thus, $K=T_B$, since the intersections of $BI$ and $(BC)$ are $B$ and $T_B$. Now, we have an orthic triangle configuration with $D,T_B,T_C$ in $\triangle HBC$. Hence, $I$ is the incenter of $\triangle DT_BT_C.$ Since $T_B$ and $T_C$ lie on $EF$, the foot from $I$ to $T_BT_C$ is just the midpoint of $EF$. Hence, in $\triangle DT_BT_C$, the midpoint of the D-altitude, $P$, the intouchpoint to $T_BT_C$, which is the midpoint of $EF$, and the $D$-excenter, which is $H$, are collinear, hence $PH$ bisects $EF$ as desired.

My sol is crazy long bruh. Just noticed that this actually connects up with 2011 ISL G4. One first considers the following claim. Claim (Partial Iran Lemma) : The feet of the perpendiculars from $B$ to $CI$ and $C$ to $BI$ respectively lie on $\overline{EF}$. Let $C_1=\overline{EF} \cap \overline{CI}$. Then, note that \begin{align*} 2 \measuredangle EC_1I &= 2 \measuredangle ECI +2 \measuredangle C_1EC\\ &= \measuredangle ACB + 2 \measuredangle FEA\\ &= \measuredangle ACB + \measuredangle FAE \\ &= \measuredangle ACB + \measuredangle BAC\\ &= \measuredangle ABC\\ &= 2\measuredangle FBI \end{align*}Thus, $\measuredangle EC_1I= \measuredangle FBI$. This means that $FBIC_1$ is cyclic and thus $\measuredangle IC_1B = 90^\circ$ which means that indeed $C_1$ is the foot of the perpendicular from $B$ to $CI$. Similarly, the foot of the perpendicular from $C$ to $BI$ also lies on $EF$ and we have our claim. Now, we have the following very important claim. Claim (Extended 11SLG4) : Consider a triangle $ABC$ with $D,E,F$ the feet of the perpendiculars from $A,B,C$ to the opposite sides. Let $H$ be its orthocenter and $Y_A$ be the $A-$Why Point of $\triangle ABC$. Let $P$ be the foot of the altitude from $D$ to $EF$ Then, the following points are collinear, 1. $A$ 2. Foot of the perpendicular from $H$ to $EF$ ($Z$) 3. Midpoint of the altitude from $D$ to $EF$ ($M$) 4. $Y_A$ First, consider the inversion centred at $A$ with radius $\sqrt{AD \cdot AH}$. This clearly maps $D$ to $H$, $B$ to $F$ and $C$ to $E$ (and vice versa). Thus, $\overline{EF}$ maps to $(ABC)$ and $\overline{BC}$ maps to $(AEF)$ and thus the $A-$Ex Point $X_A$ maps to the $A-$Queue Point $Q_A$ under this inversion. Further, since $D$ and $H$ map to each other, this in fact means that the circle $(X_AH)$ remains fixed under this inversion ($Q_A$ clearly lies on this circle as $\measuredangle HQ_AX_A=90^\circ$). Further, $Z$ will map to the intersection of $(X_AH)$ and $(ABC)$. The claim is that this intersection is $Y_A$. Let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$. Let $G$ be the centroid of $\triangle ABC$ and $W$ the intersection of $\overline{AM}$ and $(ABC)$. It is well known (from 11SLG4) that $Y_A-D-G-A'$ (with $A'G=2DG$). Simply note that \[GM \cdot GW = \frac{AG \cdot GW}{2} = \frac{A'G\cdot GY_A}{GD \cdot GY_A} = GD \cdot GY_A\]Thus, $DMWY_A$ is cyclic. Now, let $X_A' = \overline{Y_AW} \cap \overline{BC}$. Then, \[X_A'D\cdot X_A'M = X_A'Y_A\cdot X_A'W = X_A'B \cdot X_A'C\]Thus, $X_A' \equiv X_A$ and we have that $X_A-Y_A-W$. Let $H_A$ be the $A-$Humpty Point. It is well known that $X_A-H-H_A$ and $HH_AMD$ is cyclic. Now, note that from this it follows that, \[\measuredangle X_AY_AD = \measuredangle WY_AD = \measuredangle AMD=\measuredangle H_AMD = \measuredangle H_AHA = \measuredangle X_AHD\]and thus, $X_AHDY_A$ is cyclic and $Y_A$ lies on $(X_AH)$ as desired. Thus, $Z$ maps to $Y_A$ in the previously described inversion and we have that $A-Z-Y_A$. Now, we deal with the other part of the collinearity. Let $N$ be the midpoint of $X_AD$. We already know that $X_AY_ADZ$ is cyclic. Further, since $B,C,H_A$ and $Y_A$ are cyclic and $(X_AD;BC)=-1$ and we also have, \[\measuredangle H_AY_AD = \measuredangle H_AA' =90^\circ\]$Y_A$ must be the $H_A-$Humpty Point of $\triangle X_ADH_A$. Further, this implies that $H_AY_A$ is the $H_A-$median of this triangle which implies $H_A-Y_A-N$. Thus, $\measuredangle NY_AD = 90^\circ$. Further, $MN \parallel X_AP$ by Midpoint Theorem, which implies that $NY_ADM$ must indeed be cyclic. Now, \[\measuredangle DY_AM = \measuredangle DNM = \measuredangle DX_AZ = \measuredangle DY_AZ\]This clearly implies that $Y_A-M-Z$ and we put these two together and conclude that indeed $A-Z-M-Y_A$ are collinear as claimed. By the first claim, $EF$ is the line joining the feet of the altitudes from $B$ and $C$ to the opposite sides. One now notices that by applying the above-proved claim on $\triangle BHC$, we must have $H, M$ and the foot of the perpendicular from $I$ to $EF$ collinear. But, clearly $AI$ is the perpendicular bisector of $EF$ meaning that, the perpendicular from $I$ to $EF$ passes through the midpoint of $EF$. Thus, we have that $\overline{PH}$ bisects $\overline{EF}$ as was required.

Let $N$ be the midpoint of $[EF]$,$B'=(BI)\cap (EF)$and $C'= (CI)\cap(EF)$. By the Iran lemma (provable by angle chase), the triangle $\triangle DB'C'$ with incenter $I$ is the orthic triangle of $\triangle HBC$. Now since $P$ is the midpoint of the $D-$altitude in $\triangle DB'C'$ and $H$ the $D-$excenter point, the collinearity is well known(4.3 midpoints of Altitudes EGMO). $$\mathbb{Q.E.D.}$$

Let $K = BI \cap EF$ and $L = CI \cap EF$. Iran Lemma tells us $\triangle DKL$ is the orthic triangle of $\triangle HBC$. Hence Midpoint of the Altitude Lemma this triangle with altitude midpoint $P$, touch point $M$ (midpoint of $EF$), and excenter $K$ finishes. $\blacksquare$

alright here's the stock solution I guess Let $X$ and $Y$ be the feet from $C$ to $\overline{BI}$ and from $B$ to $\overline{CI}$, which both lie on $\overline{EF}$. It follows that $DXY$ is the orthic triangle of $HBC$, i.e. $H$ is the $D$-excenter of triangle $DXY$ and $I$ is the incenter. Midpoint of altitudes lemma implies $K = \overline{PH} \cap \overline{EF}$ satisfies $\angle IKE = 90^\circ$, or $K$ is the midpoint of $\overline{EF}$.

WOAH. Took a "hint" because I heard from somewhere that this is the problem after which the "Iran Lemma" is named. Let $X$ and $Y$ be the feet of the altitudes from $B,C$ to $CI,BI,$ respectively. By the Iran Lemma, we see that $X,Y,E,F$ are collinear. The condition that $PH$ bisects $EF$ is equivalent to $PH$ passing through the foot of the altitude from $I$ to $XY.$ Since $DYX$ is the orthic triangle of $\triangle BHC,$ we see that $I$ is the incenter of $\triangle XYD,$ so the foot from $I$ to $XY$ is the $D$-intouch point of $\triangle DXY.$ Additionally, it is clear that $H$ is the $D$-excenter of $\triangle DXY.$ Thus we must show that the midpoint of the $D$-altitude, the $D$-excenter, and the $D$-intouch point are collinear, which is well-known.

We use complex bash with the unit circle as the incircle. First, we calculate the circumcenter of $(BIC)$. We know that $(AB)(AC) = (AI)(AI_A)$, and by considering angles we can see that $\frac{(a - b)(a - c)}{(a - i)} = (a - i_a)$. So we have $a = \frac{2ef}{e + f}$ trivially and cyclic variants. Now we write $$\frac{(a - b)(a - c)}{(a - i)} = \frac{4ef(\frac{f}{e + f} - \frac{d}{d + e}) (\frac{e}{e + f} - \frac{d }{d + f})}{\frac{2ef}{e + f}} = \frac{2(ef - de)(ef - df)}{(e + f)} = \frac{2ef(d - f)(d - e)}{(e + f)(d+e)(d+f)}$$. Now we find $$i_a = \frac{2ef}{e + f} (1 - \frac{(d - e)(d-f)}{(d+e)(d+f)}) = \frac{2ef}{e + f} \frac{2de + 2df}{(d + e)(d+f)} = \frac{4def}{(d+e)(d+f)}$$. The circumcenter of $(BIC)$ is the midpoint of $II_a$, which is just $\frac{2def}{(d + e)(d+f)}$. Now the orthocenter is just given by $b + i + c - 2o = \frac{2de}{d + e} + \frac{2df}{d + f} - \frac{4def}{(d + e)(d + f)} = 2d(\frac{e}{d + e} + \frac{f}{d + f} - \frac{2ef}{(d + e)(d+f)}) = 2d(\frac{de + ef + df + ef - 2ef}{(d + e)(d + f)}) = \frac{2d^2(e+f)}{(d+e)(d+f)}$. Now, our strategy will be to show that the midpoint of $EF$ is collinear with $P,H$. Find the midpoint as $\frac{e + f}{2}$, now find $M$ as $\frac 12 (d + e + f - \frac{ef}{d})$, and then $P$ is given as $\frac{3d + e + f - \frac{ef}{d}}{4}$. Then we desire $\frac{2e + 2f - 4p}{e + f - 2h}$ is real, which is equivalent to showing $$\frac{e + f - 3d + \frac{ef}{d}}{e + f - 4d^2\frac{(e+f)}{(d+e)(d+f)}} = \frac{(d+e)(d+f)(e + f - 3d + \frac{ef}{d})}{((d+e)(d+f)(e+f)-4d^2(e+f))} = \frac{(d+e)(d+f)(e + f -3d + \frac{ef}{d})}{e^2f+f^2e+e^2d+f^2d + 2def - 3d^2e-3d^2f} = \frac{(d+e)(d+f)}{(de+df)}$$which is obviously self conjugating, so we are done.

Notice that in $\Delta BHC$, $I$ is the orthocenter due to Iran lemma and it is also the incenter of orthic triangle of $\Delta BHC$. The midpoint of $EF$ is the foot of altitude from $I$ to $EF$ and $P$ is the midpoint of the $D$ altitude of the orthic triangle of $BHC$ and $H$ is the $D$ excenter of this orthic triangle so the desired result follows by midpoint of altitude lemma.

Let $Q$ be the midpoint of $EF$, let $S$ be the foot of the perpendicular from $H$ to $EF$, and let $K$ be the "Iran lemma point" (the concurrency point of $EF$, $CH$, $BI$). Then \[-1=(EF\cap BC,D;C,B)\stackrel{K}{=}(RD;HI)\stackrel{\infty_{\perp EF}}{=}(RM;SQ)\stackrel{H}{=}(D,M;\infty_{DM},HQ\cap DM),\]so $HQ$ bisects $MD$, as desired.

Let $N$ be the midpoint of $EF$, and note \[(MD;P\infty) \overset{N}{=} (MN \cap HD, D; HI) = -1. \quad \blacksquare\]

Synthetic sol

$ $

Let $H_B$ and $H_C$ be the feet of the altitudes from $B$ and $C$ to $CI$ and $BI$, respectively. It's easy to see that $FH_BIDB$ and $EH_CIDC$ are cyclic. Claim: $H_B$ and $H_C$ lie on $EF$. It's clear by \[\measuredangle H_BFI=\measuredangle H_BBI=90^\circ-\measuredangle BIH_B=90^\circ-\measuredangle BIC=-\tfrac{\measuredangle BAC}{2}=\measuredangle EFI\]with $H_C$ following analogously. Let $N$ be the intersection of $HP$ with $EF$. Let $H_A$ be the foot of the altitude from $H$ to $EF$. Let $T$ be the intersection of $DH$ and $EF$. We will use LinPOP on $(BID)$ and $(CID)$. For any point $X$ let $f(X)$ be $\text{Pow}_{(BID)}(X)-\text{Pow}_{(CID)}(X)$. LinPOP states that $f$ is linear. It suffices to show that $f(N)=\tfrac{f(F)+f(E)}{2}$. Since $T$ is on the radical axis of the two circles, $f(T)=0$. By Menelaus on $PNH$ traversing $\triangle MTD$, \[\frac{MN}{TN}\cdot \frac{TH}{DH}\cdot \frac{DP}{MP}=1\implies \frac{MN}{TN}=\frac{DH}{TH}\implies \frac{NT}{MT}=\frac{TH}{DH+TH}\]Therefore, \[f(N) = \frac{TH}{DH+TH}f(M)\]Since $\angle DFE=\angle CBH$ and $\angle DEF=\angle BCH$, $\triangle DEF\sim \triangle HCB\sim \triangle HH_BH_C$. Therefore, \[\frac{TH}{DH+TH}=\frac{d(H,H_BH_C)}{2d(H,H_BH_C)+d(D,EF)}=\frac{H_BH_C}{2H_BH_C+EF}\]Let $MF=u$, $ME=v$, $MH_B=x$, $MH_C=y$. Furthermore, set $MD=1$. We have \begin{align*} 2f(N) &= \frac{2x+2y}{2x+2y+u+v} \cdot (xu-vy) \\ f(F)+f(E) &= (u+v)(v+x-u-y) \end{align*}Now, we relate $x$ with $v$ and $y$ with $u$. Claim: $2x=\tfrac{1}{v}-v$; $2y=\tfrac{1}{u}-u$. It is simple. We have by orthocenter-incenter duality that $\triangle DH_BH_C\sim \triangle ABC$ implying $\angle DH_CM=\angle C$ while $\angle MDF=90^\circ-\angle EFD=\tfrac{\angle C}{2}$ so \[y=\cot{\angle C}=\frac{1-\tan^2\left(\frac{\angle C}{2}\right)}{2\tan\left(\frac{\angle C}{2}\right)} = \frac{1-u^2}{2u}\]and the result follows. Therefore, we have \begin{align*} 2f(N) &= \left(\frac{\frac{1}{v}-u+\frac{1}{u}-v}{\frac{1}{u}+\frac{1}{v}}\right) \left(xu-vy\right) \\ &= \frac{1}{2}\left(\frac{u+v-uv(u+v)}{u+v}\right) \left(\frac{u}{v}-\frac{v}{u}\right) \\ &= \frac{1}{2}\left(\frac{u+v-uv(u+v)}{u+v}\right) \left(\frac{(u-v)(u+v)}{uv}\right) \\ &= \frac{1}{2}\left(\frac{(u+v-uv(u+v))(u-v)}{uv}\right) \\ &= (u+v)\left(\frac{(1-uv)(u-v)}{2uv}\right) \\ &= (u+v)\left(\frac{u+uv^2}{2uv}-\frac{v+u^2v}{2uv}\right) \\ &= (u+v)\left(\frac{1+v^2}{2v}-\frac{1+u^2}{2u}\right) \\ &= (u+v)\left(\frac{1-v^2}{2v}+v-\frac{1-u^2}{2u}-u\right) \\ &= (u+v)\left(x+v-y-u\right) \end{align*}and we are done.