Ok, here's a quick sketch: If $M=CH_2\cap AB,N=BH_2\cap CD$, show that $MN\|AD$. From here, we find that $MH_2\|AH_1,NH_2\|DH_1,MN\|AD$, so $AH_1D,MH_2N$ are homothetic, meaning that $AM,DN,H_1H_2$ are concurrent.

Stun wrote: Suppose that $ABCD$ is a cyclic quadrilateral. Let $E=AC\cap BD$ and $F=AB\cap CD$. Denote by $H_{1}$ and $H_{2}$ the orthocenters of triangles $EAD$ and $EBC$, respectively. Prove that the points $F$, $H_{1}$, $H_{2}$ are collinear. I am amused that this problem appeared as G8 in a recent shortlist. It's absolutely trivial. In fact, it is well-known ( http://cut-the-knot.com/Curriculum/Geometry/CircleOnCevian.shtml , theorem 1) that, if we have a triangle and two cevians of this triangle (issuing from different vertices), and consider the circles with these cevians as diameters, then the orthocenter of the triangle lies on the radical axis of these two circles. Applying this fact to the triangle EAD with the orthocenter $H_{1}$ and the two cevians AB and CD, it follows that the orthocenter $H_{1}$ of the triangle EAD lies on the radical axis of the circles with diameters AB and CD. Applying the same fact to the triangle EBC with the orthocenter $H_{2}$ and the two cevians AB and CD, we see that the orthocenter $H_{2}$ of the triangle EBC lies on the radical axis of the circles with diameters AB and CD. Finally, the point F also lies on the radical axis of the circles with diameters AB and CD, since the power of the point F with respect to the circle with diameter AB equals $FA\cdot FB$, and the power of the point F with respect to the circle with diameter CD equals $FC\cdot FD$, but we have $FA\cdot FB=FC\cdot FD$ by the intersecting chords theorem, since the quadrilateral ABCD is cyclic. Thus, all three points F, $H_{1}$ and $H_{2}$ lie on the radical axis of the circles with diameters AB and CD; thus, these three points are collinear. $\blacksquare$ darij

I have observed many solutions part however part I thought did not see any solution.Then let's solve the second part: M and N are the cut-off points BH with CC´ and CH with BB´ BH and C´H´ are perpendicular to BC = > BH / / C´H´ mode analogous CH / / B´H´, also know that < ABH = 90 - < BAC = < ACH as Quadrilatero BC´B´C is ciclico = > < ABB´ = < ACC´ so get that < NBM = < MCN this implies that the BNMC Tomko is ciclico then < NBC = < C´MN = < MC´B´, MN and C´B´ lines are parallel. Therefore C´H´B´ and NHM triangles are so homotéticos C´M, B´N and HH´ lines are concurrent i.e. equivalent lol CC´, BB´ and HH´ are concurrent.

luca-97 wrote: I have observed many solutions part however part I thought did not see any solution. That's because the two "parts" are equivalent. The points $A$, $B$, $C$, $D$, $E$, $F$, $H_1$ and $H_2$ of the first part are the points $B$, $B'$, $C'$, $C$, $A$, $BB'\cap CC'$, $H$ and $H'$ of the second part, respectively.

Is there another solutions??

Is there another solutions??

Dear Mathlinkers, see also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=497762 Sincerely Jean-Louis

See a generalization here

Dear Mathlinkers, in my point of view, this problem can be seen as a generalization of a result of von Nagel : AO and AH being two A-sogonal line of ABC, O, H the circumcenter, orthocenter... which leads to a proof, before a new generalization. Sincerely Jean-Louis

this is just the coaxial line of the diametre circles of $AB, CD$... obviously $F$ is on it.

This Problem is a straightforward application of the Gauss-Bodenmiller theorem to the quadrilateral QAEB, where Q is the inetersection of AD and BC.

Dear Mathlinkers, have also a look at http://jl.ayme.pagesperso-orange.fr/Docs/von%20Nagel%20revisited%20and%20generalized.pdf Sincerely Jean-Louis

Did I made any mistake or this seems way too easy for a G8 ? Construct two circles $\omega_1$ and $\omega_2$ with diameters $\overline{AB}, \overline{CD}$ respectively. Let $\ell$ be the radical axis of $\omega_1, \omega_2$. Since $$\text{Pow}_F(\omega_1) = FA \times FB = \text{Pow}_F(\omega_{ABCD}) = FC \times FD = \text{Pow}_F(\omega_2)$$, we have $F \in \ell$ Let $H_1$ be the orthocentre of $\Delta EAD$, and let $P_A, P_D$ be the perpendicular from $H_1$ to $ED, EA$ respectively. Since $\angle AP_AD = 90 = \angle AP_DD$, we have $AP_DP_AD$ concyclic. On the other hand, as $\angle BP_AA = 180 - \angle AP_AD = 90$, we have $P_A \in \omega_1$ and similarly, $P_D \in \omega_2$. Now $$\text{Pow}_{H_1}(\omega_1) = H_1A \times H_1P_A = \text{Pow}_{H_1}(\omega_{AP_DP_AD}) = H_1P_D \times HD = \text{Pow}_{H_1}(\omega_2)$$, and hence $H_1 \in \ell$ Let $H_2$ be the orthocentre of $\Delta EBC$. By exact same reasoning as above, we see $H_2 \in \ell$ too. As all three points lie on the line $\ell$, the three points are colinear as desired.

Citing the same lemma for second time in a day. The problem is straight "Parallelogram Isogonality Lemma", a special case of the well known Isogonality Lemma.

Consider $(BB')$ , $(CC')$ and the circle $\omega$ . If $ P = B'B\cap CC'$ , then , P is the radical center of the three circles .... Consider $\Delta ABC$ , then $BB'$ and $CC'$ are cevians and hence $H$ lies on radical axis of $(BB')$ , $(CC')$ . looking w.r.t $\Delta AB'C'$ , $H'$ also lies on this line , so done

İsn’t it Gauss-BodenmillerTheorem?

Let $\omega_1$ and $\omega_2$ denote the circles with diameter $AB$ and $CD$. It suffices to show that $F, H_1, H_2$ lie on the radical axis of $\omega_1$ and $\omega_2$. For $F$, notice that it is the radical center of $(ABCD), \omega_1, \omega_2$, so $F$ is on the radical axis of $\omega_1$ and $\omega_2$. Now, notice that the perpendicular from $A$ to $DE$ is on $\omega_1$, and the perpendicular from $D$ to $AE$ is on $\omega_2$. Meanwhile, the points $A, D$, the foot from $A$ to $DE$ and the foot from $D$ to $AE$ are concyclic (call this circle $\Gamma$), so $H_1$ is the radical center of $\omega_1, \omega_2$, and the circle $\Gamma$. This means that $H_1$ lies on the radical axis of $\omega_1$ and $\omega_2$. Similar reasoning can be used to show that $H_2$ lies on the radical axis of $\omega_1$ and $\omega_2$, so $F, H_1, H_2$ all lie on the radical axis and they're collinear.

Let $H_1=X,H_2=Y$, and $Z=AX \cap BY, W= DX \cap CY, T=BX \cap CY, U= CX \cap DY$. Clearly, $\angle EXD=90º- \angle ADB= 90º- \angle ACB= \angle YEC$, so $EX,EY$ are isogonal WRT $\angle BEC$. Clearly, $W,Z$ are the ortocenters of $ECD,EAB$, respectively. Also, by the Isogonality Lemma WRT $(EC,ED),(EX,EY)$, we have that $EW,EU$ are isogonal WRT $\angle CED$. Thus, since $EW \perp CD$, $EU$ passes through the circumcenter of $ECD$, but since $EDC ~ EAB$, $EW$ passes through the circumcenter of $EAB$. Similarly, we also have that $ET$ passes through the circumcenter of $EAB$, so $E,W,T$ are collinear. Now, since $BX \cap AY= T, DX \cap CY= W, AC \cap BD= E$ are collinear, by Desargues' Theorem, we have that $AB,XY,CD$ are concurrent, so $F$ lies on $XY$, as desired. $\blacksquare$

Solved with DebayuRMO, Fakesolver19 and carried by another guy from discord (whom I'm not exactly sure how to credit) because we were busy finding "DegEnErAtE vErSiOn oF sTeInEr". Let $\omega_1$ and $\omega_2$ denote the circles with diameter $AB$ and $CD$ respectively. By radical axis theorem on $(\omega_1 , (ABCD) , \omega_2)$ we see that $F$ lies in the radical axis of $(\omega_1 , \omega_2)$. Let $X$ denote the foot of perpendicular from $A$ to $DE$ and $Y$ denote the foot of perpendicular from $D$ to $AE$, also let $H_1$ denote the orthocenter of $\triangle ADE$ and $H_2$ denote the orthocenter of $\triangle BCE$. Note that $X \in \omega_1$ and $Y \in \omega_2$. $$\text{Pow}(H_1 , \omega_1) = AH_1 \cdot H_1X = DH_1 \cdot H_1Y = \text{Pow}(H_1 , \omega_2)$$So $H_1$ lies in the radical axis of $(\omega_1 , \omega_2)$, similarly $H_2$ lies in the radical axis of $(\omega_1 , \omega_2)$ thus we get that $\overline{F-H_1 - H_2}$ are collinear because all of them lie on the mentioned radical axis.

Let $J = AC \cap BH_2 , I = BD \cap AH_1 , K = AC \cap DH_1 , L = BD \cap CH_2$. It is clear that $AJIB , DLKC , JLKI$ are concyclic. Let $G = JI \cap KL$. Clearly $G$ lies on the radical axis of $(AJIB)$ and $(DLKC)$. Since also Pow($J , (AJIB)) = $Pow($J , (DLKC))$, $GF$ is the radical axis of $(AJIB)$ and $(DLKC)$. But, since $H_1I \cdot H_1A = H_1K \cdot H_1D,$ $H_1$ also lies on this radical axis. Similarly, $H_2J \cdot H_2B = H_2L \cdot H_2C$, implying that $H_2$ also lies on this radical axis. So, $F , H_1, H_2 , G$ are collinear, as desired.

MrOreoJuice wrote: Solved with DebayuRMO, Fakesolver19 and carried by another guy from discord (whom I'm not exactly sure how to credit) because we were busy finding "DegEnErAtE vErSiOn oF sTeInEr". Let $\omega_1$ and $\omega_2$ denote the circles with diameter $AB$ and $CD$ respectively. By radical axis theorem on $(\omega_1 , (ABCD) , \omega_2)$ we see that $F$ lies in the radical axis of $(\omega_1 , \omega_2)$. Let $X$ denote the foot of perpendicular from $A$ to $DE$ and $Y$ denote the foot of perpendicular from $D$ to $AE$, also let $H_1$ denote the orthocenter of $\triangle ADE$ and $H_2$ denote the orthocenter of $\triangle BCE$. Note that $X \in \omega_1$ and $Y \in \omega_2$. $$\text{Pow}(H_1 , \omega_1) = AH \cdot HX = BH \cdot HY = \text{Pow}(H_1 , \omega_2)$$So $H_1$ lies in the radical axis of $(\omega_1 , \omega_2)$, similarly $H_2$ lies in the radical axis of $(\omega_1 , \omega_2)$ thus we get that $\overline{F-H_1 - H_2}$ are collinear because all of them lie on the mentioned radical axis. 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The proof of $F$ lying on the radical axis is essentially the same as above and $H_1$ and $H_2$ lie on the radical axis from the proof of Steiner line.

Trivial G8. Let $\omega_1$ be the circle with diameter $AB$ and let $\omega_2$ be the circle with diameter $CD$. Let $CC_1$ and $BB_1$ be the altitudes of $EBC$ and let $AA_1$ and $DD_1$ be the altitude of $EAD$. Note that $A_1,B_1$ lie on $\omega_1$ and $C_1,D_1$ lie on $\omega_2$. Now as $H_1A\cdot H_1A_1=H_1D\cdot H_1D_1$, we have that $H_1$ lies on the radical axis of $\omega_1$ and $\omega_2$, similarly $H_2$ lies on the radical axis of $\omega_1$ and $\omega_2$. Hence by the radical axis theorem on $\omega_1,\omega_2$ and $(ABCD)$, we get that $AB,CD,H_1H_2$ are concurrent, we are done.

Collinearity associated with cyclic quads is almost always an application of radical axes!

Let $H_1$ and $H_2$ denote the orthocenters of $EBC$ and $EAD$ respectively, $Q, R$ be the projections of $H_1$ onto $AC, BD$ respectively, and $Y, Z$ be the projections of $H_2$ onto $BD, AC$ respectively. By right angles, it's easy to see $AYQB$ and $CRZD$ are cyclic. Let the circumcircles of these two cyclic quadrilaterals be $\omega_1$ and $\omega_2$ respectively. Claim: $F, H_1, H_2$ each lie on the Radical Axis of $\omega_1$ and $\omega_2$. Proof. Notice $$Pow_{\omega_1}(F) = FA \cdot FB = Pow_{(ABCD)}(F) = FC \cdot FD = Pow_{\omega_2}(F);$$$$ Pow_{\omega_1}(H_1) = H_1B \cdot H_1Q = Pow_{(BCQR)}(H_1) = H_1C \cdot H_1R = Pow_{\omega_2}(H_1);$$$$Pow_{\omega_1}(H_2) = H_2A \cdot H_2Y = Pow_{(ADYZ)}(H_2) = H_2D \cdot H_2Z = Pow_{\omega_2}(H_2).$$$\square$ This directly implies the desired result. $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.1; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(2); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.6412591300876, xmax = 4.776744629047886, ymin = -4.432815355655552, ymax = 5.801392669802317; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccwwff = rgb(0.8,0.4,1); pen ffcctt = rgb(1,0.8,0.2); pen zzffff = rgb(0.6,1,1); pen ffttww = rgb(1,0.2,0.4); /* draw figures */ draw(circle((-0.5,1.0241089245354051), 3.8528079122183043), linewidth(0.4) + blue); draw((xmin, 2.5894919081674974*xmin + 9.771446615706738)--(xmax, 2.5894919081674974*xmax + 9.771446615706738), linewidth(0.4)); /* line */ draw((xmin, -6.510570233503604*xmin + 18.945189683547564)--(xmax, -6.510570233503604*xmax + 18.945189683547564), linewidth(0.4)); /* line */ draw((-2.009020903170895,4.569103243606349)--(2.355293051173564,3.6108888533990737), linewidth(0.4)); draw((-4,-0.5865210169632515)--(3,-0.5865210169632515), linewidth(0.4)); draw((3,-0.5865210169632515)--(-2.009020903170895,4.569103243606349), linewidth(0.4)); draw((-4,-0.5865210169632515)--(2.355293051173564,3.6108888533990737), linewidth(0.4)); draw((xmin, 4.554632031147433*xmin + 1.0275291486590357)--(xmax, 4.554632031147433*xmax + 1.0275291486590357), linewidth(0.4) + dotted + qqwuqq); /* line */ draw((xmin, -1.5140987531496344*xmin + 1.5272471990637446)--(xmax, -1.5140987531496344*xmax + 1.5272471990637446), linewidth(0.4) + dotted + qqwuqq); /* line */ draw((xmin, 0.9715643828973471*xmin + 1.3225700135932201)--(xmax, 0.9715643828973471*xmax + 1.3225700135932201), linewidth(0.4) + dotted + qqwuqq); /* line */ draw(shift((-3.0045104515854475,1.9912911133215485))*xscale(2.7633520984568847)*yscale(2.7633520984568847)*arc((0,0),1,-111.11537135625291,68.88462864374709), linewidth(0.4) + ccwwff); draw(shift((-3.0045104515854475,1.9912911133215485))*xscale(2.7633520984568847)*yscale(2.7633520984568847)*arc((0,0),1,68.8846286437471,248.88462864374708), linewidth(0.4) + ccwwff); draw(shift((2.6776465255867823,1.512183918217911))*xscale(2.1233167845189844)*yscale(2.1233167845189844)*arc((0,0),1,-81.26781852784436,98.73218147215566), linewidth(0.4) + ffcctt); draw(shift((2.6776465255867823,1.512183918217911))*xscale(2.1233167845189844)*yscale(2.1233167845189844)*arc((0,0),1,98.73218147215567,278.7321814721557), linewidth(0.4) + ffcctt); draw((xmin, 11.859886545730326*xmin + 0.4259919123750597)--(xmax, 11.859886545730326*xmax + 0.4259919123750597), linewidth(0.4) + linetype("4 4")); /* line */ draw(shift((1.309611054155772,2.9202588698246466))*xscale(1.2531642402730505)*yscale(1.2531642402730505)*arc((0,0),1,33.44312241132928,213.44312241132926), linewidth(0.4) + zzffff); draw(shift((1.309611054155772,2.9202588698246466))*xscale(1.25316424027305)*yscale(1.25316424027305)*arc((0,0),1,-146.55687758867074,33.44312241132926), linewidth(0.4) + zzffff); draw(shift((-0.8725459230164576,3.399366064928284))*xscale(1.6309078599659599)*yscale(1.6309078599659599)*arc((0,0),1,134.17368770457352,314.1736877045735), linewidth(0.4) + ffttww); draw(shift((-0.8725459230164576,3.399366064928284))*xscale(1.6309078599659599)*yscale(1.6309078599659599)*arc((0,0),1,-45.82631229542648,134.1736877045735), linewidth(0.4) + ffttww); /* dots and labels */ dot((-2.009020903170895,4.569103243606349),linewidth(2pt) + dotstyle); label("$A$", (-1.9495115889416594,4.626711559962027), NE * labelscalefactor); dot((-4,-0.5865210169632515),linewidth(2pt) + dotstyle); label("$B$", (-3.928768801412267,-0.5225754968721207), NE * labelscalefactor); dot((3,-0.5865210169632515),linewidth(2pt) + dotstyle); label("$C$", (3.071043291471589,-0.5225754968721207), NE * labelscalefactor); dot((2.355293051173564,3.6108888533990737),linewidth(2pt) + dotstyle); label("$D$", (2.4273824093673264,3.677311758858231), NE * labelscalefactor); dot((0.26392905713797987,2.22962888625022),linewidth(2pt) + dotstyle); label("$E$", (0.33548454252847276,2.293440862334054), N * labelscalefactor); dot((1.008096749782875,12.381904991919448),linewidth(2pt) + dotstyle); label("$F$", (0.8665047702644895,5.543928316960609), NE * labelscalefactor); dot((0.08234309086468902,1.402571627855032),linewidth(2pt) + dotstyle); label("$H_1$", (0.14238627789719396,1.4727732376511116), NE * labelscalefactor); dot((0.26392905713797987,3.5561605861530774),linewidth(2pt) + dotstyle); label("$H_2$", (0.33548454252847276,3.6129456706478043), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $H_1$ and $H_2$ be the orthocenters of $\triangle EAD$ and $\triangle EBC,$ respectively, and $\omega_1,\omega_2,\omega_3,\omega_4$ be the circles with diameters $\overline{AB},\overline{CD},\overline{AE},\overline{DE},$ respectively. Notice that $H_1$ lies on the radical axis of $\omega_2$ and $\omega_4$ since $$\angle D(\overline{DH_1}\cap\overline{AC})C=90.$$Similarly, $H_1$ lies on the radical axis of $\omega_1$ and $\omega_3.$ Finally, $H_1$ lies on the radical axis of $\omega_3$ and $\omega_4$ as $$\angle E(\overline{EH_1}\cap\overline{AD})D=90.$$Hence, $$\text{pow}_{\omega_1}H_1=\text{pow}_{\omega_3}H_1=\text{pow}_{\omega_4}H_1=\text{pow}_{\omega_2}H_1$$and $H_1$ lies on the radical axis of $\omega_1$ and $\omega_2.$ Similarly, $H_2$ lies on the radical axis of $\omega_1$ and $\omega_2.$ But $F$ is the radical center of $(ABCD),\omega_1,\omega_2,$ so $F$ must also lie on the radical axis of $\omega_1$ and $\omega_2.$ $\square$

By the Gauss-Bodenmiller Theorem, if we consider complete quadrilateral $ACBD$, and let $G = AD \cap BC$, the orthocenters of $GAC, GBD, EAD, EBC$ lie on the Radical Axis of the circles with diameter $AB$ and $CD$. So it suffices to prove this radical axis hits $F$, which is true by the Radical Axis Theorem on $(ABCD)$ and these two circles. $\blacksquare$

Let $H_1$ be the orthocenter of $\triangle EAD$, $H_2$ the orthocenter of $\triangle EBC$. As $\triangle EAD\sim \triangle EBC$, this means $\triangle AH_1D\sim \triangle BH_2C$. Let $H_1'$ denote the reflection of $H_1$ over the midpoint of line $AD$. It is clear that $FBH_2C\sim FDH_1'A$ the angles add up properly and $\triangle FBC\sim \triangle FDA$, so $FH_1'$ is isogonal to $FH_2$. But by the first isogonality lemma, $FH_1$ is isogonal to $FH_1'$, so $FH_1$ and $FH_2$ are the same line as desired.

Let $\mathcal{F}$ be the parabola touching to sides of complete quadrilateral $\left\{ AD,DB,BC,CA\right\} ;$ $\ell =H_1H_2$ is the directrix. We consider dual transformation $\gamma$ wrt $\mathcal {F};$ $\gamma (\ell)$ is focus of parabola and Miquel point of complete quadrilateral, so by radical axis lines $\overline{\gamma (\ell)E}, AD,BC$ concur, or equivalently $\gamma (AD),$ $\gamma (BC),$ $\ell \cap \overline{\gamma (AC)\gamma (BD)}$ are collinear. But since $\overline{\gamma (AD)\gamma (BC)} ,\overline{\gamma (AC)\gamma (BD)}$ concur on $\ell,$ points $$\overline{\gamma (AD)\gamma (BD)} \cap \overline{\gamma (AC)\gamma (BC)} ,\overline{\gamma (AD)\gamma (AC)} \cap \overline{\gamma (BD)\gamma (BC)}$$are collinear with $\gamma (\ell)$ or equivalently $\ell ,AB,CD$ concur.

We use the alternative formulation. Let $X = \overline{BB'} \cap \overline{HH'}$. Our goal is to show that $\overline{BB'}$ and $\overline{CC'}$ bisect $H$ with the same ratio. Write $$\frac{H'X}{\sin \angle HB'B} = \frac{H'B'}{\sin \angle H'XB'} \text{ and } \frac{HX}{\sin \angle BB'H} = \frac{HB}{\sin \angle HXB},$$and thus $$\frac{H'X}{HX} = k \cdot \frac{\sin \angle HB'B}{\sin \angle BB'H}$$where $k$ is the similarity ratio. Now, as $$\angle HB'B=90^\circ -C+\angle C'CB \text{ and } \angle HBB' = 90^\circ - A - \angle C'CA,$$we have $$\sin \angle HB'B = \sin(90^\circ - \angle C'CA) = \cos \angle C'CA$$and $\sin \angle HB'B = -\cos \angle AC'C$. But $\angle AC'C = \angle AB'B$ and $\angle CC'A = \angle BB'A$, so this expression is symmetric in $B$ and $C$, and we are done.

Consider the circle $\omega_1$ with diameter $\overline{AB}$ and the circle $\omega_2$ with diameter $\overline{CD}$. Moreover, let $\omega$ be the circumcircle of $ABCD$. [asy][asy] size(5cm); pair A = dir(120); pair D = dir(90); pair B = dir(210); pair C = dir(180)/B; draw(unitcircle); draw(A--B--C--D--cycle); pair F = extension(A, B, C, D); pair E = extension(A, C, B, D); pair H_1 = orthocenter(E, A, D); pair H_2 = orthocenter(E, B, C); dot(H_1); dot(H_2); draw(CP(midpoint(A--B), A), dotted); draw(CP(midpoint(C--D), C), dotted); draw(C--A--F--D--B); draw(F--(3*H_1-2*F)); dot("$A$", A, dir(A)); dot("$D$", D, dir(D)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$F$", F, dir(F)); dot("$E$", E, 1.3*dir(0)); /* Source generated by TSQ */ [/asy][/asy] We saw already in the proof of the Gauss line that the two orthocenters lie on the radical axis of $\omega_1$ and $\omega_2$ (i.e., the Steiner line of $ADBC$). Hence the problem is solved if we can prove that $F$ also lies on this radical axis. But this follows from the fact that $F$ is actually the radical center of circles $\omega_1$, $\omega_2$ and $\omega$.

ok solution 1 (smart): homothety solution 2 (also smart): radical axis/radical center trick/steiner configuration solution 3 (my solution which is really stupid): step 1: trig ceva! \[\frac{\sin \angle BAH_{AD}}{\sin \angle BAD}\cdot \frac{\sin \angle H_{BC}H_{AD}D}{\sin \angle H_{BC}H_{AD}A}\cdot \frac{\sin \angle CDA}{\sin \angle CDH_{AD}}=1\] step 2: cancel $\angle BAH_{AD}$ and $\angle CDH_{AD}$ to get: \[\frac{\sin \angle CDA}{\sin \angle BAD}\cdot \frac{\sin \angle H_{BC}H_{AD}D}{\sin \angle H_{BC}H_{AD}A}=1\] step 3: let $G=AD\cap BC$ and angle chase to get $\sin \angle GEB=\sin \angle H_{BC}H_{AD}D$ \[\frac{\sin \angle GEB}{\sin \angle GEC}\cdot \frac{\sin \angle CDA}{\sin \angle BAD}=1\] step 4: ratio lemma on $\triangle EBC$ \[\frac{GB\cdot EC\cdot FC}{GC\cdot EB\cdot FB}=1\] step 5: Menelaus on $\triangle FBC$ \[\frac{GC}{GB}\cdot \frac{AB}{AF}\cdot \frac{DF}{DC}=1\] step 6: multiply steps 4 and 5, it suffices to show that \[\frac{AB\cdot DF\cdot EC\cdot FC}{AF\cdot DC\cdot EB\cdot FB}=1\]but since \[\frac{DF\cdot FC}{AF\cdot FB}=1\]\[\frac{AB}{EB}=\frac{DC}{EC}\]we are done. (moral of the story: length bash until it works)

Let $\omega$ denote $(ABCD)$, $\omega_1$ be the circle with diameter $AB$, and $\omega_2$ be the circle with diameter $CD$. Note that the radical axis of $\omega$ and $\omega_1$ is $\overline{AB}$ and that the radical axis of $\omega$ and $\omega_2$ is $\overline{CD}$. I contend that the radical axis of $\omega_1$ and $\omega_2$ is $\overline{H_1H_2}$. Let $B_1 = \overline{B_1H_1} \cap \overline{AC}$ and $C_1 = \overline{C_1H_1} \cap \overline{BD}$. Note that \[ \text{Pow}(H_1, \omega_1) = H_1B_1 \cdot H_1B = H_1C_1 \cdot H_1C = \text{Pow}(H_1, \omega_2) \]by PoP on $H_1$ with respect to the circle with diameter $BC$, so $H_1$ lies on the radical axis $\ell$ of $\omega_1$ and $\omega_2$. By similar logic, $H_2$ also lies on $\ell$. We conclude the collinearity of $H_1$, $H_2$, and $F$ by the radical axis theorem on $\omega$, $\omega_1$, and $\omega_2$.

Muhammetnazar wrote: Is there another solutions?? Yes, a very good one We proceed with Complex Numbers Coordinates. Set the circumcircle of ABCD be the unit circle, and define $A=a,B=b,C=c,D=d$. The calculation is not difficult as a lot is symmetric and do not need to be repeated In the end you get the answer Weird that there is no other Analytic solns

@above how did you get the formulae for the orthocenters of triangles $ EAD$ and $ EBC$?

Stun wrote: Let $ ABC$ be a triangle. A circle passing through $ B$ and $ C$ intersects the sides $ AB$ and $ AC$ again at $ C'$ and $ B',$ respectively. Prove that $ BB'$, $CC'$ and $ HH'$ are concurrent, where $ H$ and $ H'$ are the orthocentres of triangles $ ABC$ and $ AB'C'$ respectively. We make quick work of this by moving points! Animate a line $\ell$ anti-parallel to $BC$ in angle $A$ with constant velocity, and let it meet $AB, AC$ at $C’, B’$ respectively. Since $\triangle AB’C’$ has fixed shape, $B’, C’, H’$ all move linearly. Thus, each of lines $BB’, CC’, HH’$ are degree $1$ and so their concurrence needs to only be verified for $4$ choices of $\ell$. When $\ell$ passes through $A$, when $\ell$ passes through feet of perpendiculars from $B, C$ to $AC, AB$; when $\ell$ passes through the intersections of perpendiculars to $AB$ at $B$ and $AC$ at $C$ with sides $AC, AB$ respectively, and finally when $\ell$ passes through the foot of the $A$ angle bisector on $BC$ — all four of these cases are degenerate and our problem is solved!

Denote by $X = AD \cap BC$. Then the orthocenters of $\triangle ADE, \triangle BEC, \triangle ACX, \triangle BXD$ are collinear through the Steiner line of quadrilateral $DECX$. Now it is well known that the circles of diameter $XE, DC, AB$ are coaxial and their radical axis is the Steiner line. Now observe that as $ABCD$ is cyclic, we have $FD \cdot FC = FA \cdot FB$, so that $F$ lies on this radical axis, as desired. $\blacksquare$

Consider the Steiner line of self-intersecting quadrilateral $ACBD$, which is the radical axis of circles $(AB)$ and $(CD)$ which we will call $\ell$. It is well known that $\ell$ passes through $H_1$ and $H_2$. And since $ABCD$ is cyclic we have $FA \cdot FB = FC \cdot FD \implies F \in \ell$, so we are done.

Clearly $F,H_1,H_2$ all lie on the radical axis of $(AB)$ and $(CD)$, so we are done.