Let E be the midpoint of AC, G of OP. angle OFP = angle OEP = 90 OFPG- inscribed in circle with center G Let K be midpiont of OH. It is obvious that K is the center ot the Euler's (also known as nine-point) circle for the triangle ABC. Than K, G lie on the perpendicular bisector of the common chord FE. angle FHP = 90- angle EFH and angle EFH = angle EFC = angle ECF = 90- angle A angle FHP = angle A There you go! Irina P.S. I think I've seen this somewhere before... Was it an IMO problem? I'm too lazy to check...

It's from a shortlist (can't remember which one); I don't know if it was actually given at an IMO, but I doubt it. Nice soln! The problem I was referring to (the one I said you could use in order to prove this) is the Butterfly problem. Here's my soln: Let T be the intersection between the altitude CH and the circumcircle of ABC. Let the chord FP (a chord in the circumcircle of ABC) cut the chord BT at Q. OF perpendicular to chord PF and O is the center of the circumcircle => F is the midpt of the chord PF and, because of the butterfly property, F must be the midpt of PQ (*). It's well-known that F is the midpt of HT (**). From (*) and (**) we get triangles FHP and FTQ equal, so HP || TQ=TB, so angle FHP=angle FTB=angle BAC Q.E.D.

u find it in our olympiad (14-th Iranian Mathematical Olympiad 1996/1997 (1375)september). it's beatifuly solved by batterfly theorem.

If somebody is still interested, I have another solution: I use the orthologic triangles theorem, which states that if ABC and A'B'C' are two non-degenerate triangles, then the lines $A\;\overline{B^{\prime }C^{\prime }}$, $B\;\overline{C^{\prime }A^{\prime }}$, $C\;\overline{A^{\prime }B^{\prime }}$ concur if and only if the lines $A^{\prime }\;\overline{BC}$, $B^{\prime }\;\overline{CA}$, $C^{\prime }\;\overline{AB}$ concur. Hereby, for any point P and any line g, the notion $P\;\overline{g}$ means the perpendicular from the point P to the line g. For your problem, I will work with directed angles modulo 180°, and I will prove that < FHP = < CAB. Let C' be the reflection of the point C in the line AB, or, equivalently, the reflection of the point C in the point F. Let also Z be the reflection of the point C in the point O. Then, the segment CZ is a diameter of the circumcircle of triangle ABC; hence, < CAZ = 90°, and thus $ZA \perp AC$. Similarly, $ZB \perp BC$. Since the points O and F are the midpoints of the segments CZ and CC', we have OF || C'Z. Now, since the point C' is the reflection of the point C in the line AB, we have < CAB = < BAC'. Thus, instead of proving < FHP = < CAB, it will be enough to show < FHP = < BAC'. But < FHP = < (FH; HP) = < (FH; AB) + < (AB; HP) = 90° + < (AB; HP), and < BAC' = < (AB; AC'). So we have to prove that 90° + < (AB; HP) = < (AB; AC'). This is equivalent to 90° = < (AB; AC') - < (AB; HP), what is obviously equivalent to 90° = < (HP; AC'). Thus, we must show that 90° = < (HP; AC'), i. e. we must show that $HP \perp AC^{\prime}$. In other words, we must show that the point P lies on the line $H\;\overline{AC^{\prime }}$. Now, the point P is defined as the point of intersection of the lines $F\;\overline{OF}$ and AC. Since OF || C'Z, we can rewrite $F\;\overline{OF}$ as $F\;\overline{C^{\prime }Z}$, and since $ZA \perp AC$, we can rewrite AC as $A\;\overline{ZA}$. Thus, we must prove that the point P, defined as the point of intersection of the lines $F\;\overline{C^{\prime }Z}$ and $A\;\overline{ZA}$, lies on the line $H\;\overline{AC^{\prime }}$. Or, simply, we have to prove that the lines $F\;\overline{C^{\prime }Z}$, $A\;\overline{ZA}$, $H\;\overline{AC^{\prime }}$ concur. By the orthologic triangles theorem, applied to the triangles FAH and AC'Z, this is equivalent to proving that the lines $A\;\overline{AH}$, $C^{\prime }\;\overline{HF}$, $Z\;\overline{FA}$ concur. In order to prove this, we denote by S the point of intersection of the lines $A\;\overline{AH}$ and $C^{\prime }\;\overline{HF}$, and try to show that this point S lies on the line $Z\;\overline{FA}$, i. e. that we have $ZS\perp FA$. Well, since the point S lies on the line $A\;\overline{AH}$, we have $AS \perp AH$, and together with $AH \perp BC$, this gives AS || BC. Since the point S lies on the line $C^{\prime }\;\overline{HF}$, we have $C^{\prime } S \perp HF$, and since $HF \perp AB$, this yields C'S || AB. If the lines CS and AB meet at K, then from C'S || AB, we have CK : KS = CF : FC', and since CF : FC' = 1 (the point C' is the reflection of the point C in the point F), we have CK : KS = 1, too, so that the point K is the midpoint of the segment CS. On the other hand, AS || BC yields BK : KA = CK : KS, what now shows us that BK : KA = 1, and the point K is the midpoint of the segment AB. Thus, the segments AB and CS have the point K as their common midpoint, i. e. these segments bisect each other, and it follows that the quadrilateral ACBS is a parallelogram. Hence, not only AS || BC, but also BS || AC. Now, BS || AC together with $ZA \perp AC$ yields $ZA \perp BS$, while AS || BC together with $ZB \perp BC$ yields $ZB \perp AS$. Hence, the point Z lies on two of the three altitudes of the triangle ABS; this means that the point Z is the orthocenter of this triangle, and hence also lies on the third altitude. And this yields $ZS \perp AB$, or, in other words, $ZS \perp FA$. Proof complete. Well, this is a really monstrous solution, but it doesn't use the butterfly theorem, does it? Darij

Have a look at page 27/52.

See the problem $P3$ from http://www.mathlinks.ro/Forum/viewtopic.php?t=46146

Let the altitude CF meet the circumcircle (O) of the triangle $\triangle ABC$ again at a point D and consider the cyclic quadrilateral ADBC with the diagonal intersection F. Let the perpendicular to OF at F meet AC at P, DB at P', the circumcircle arc DC opposite to the vertex B at X, and the circumcircle arc DC opposite to the vertex A at X'. Since $XX' \perp OF$, FX = FX'. By the butterfly theorem, FP = FP' as well, i.e., P' is a reflection of P in the line OF. Reflect the cyclic quadrilateral ADBC in the line OF into a cyclic quadrilateral A'D'B'C' with the same circumcircle (O) and the same diagonal intersection F. Then D'B' meets AC at P and A'C' meets DB at P'. (This is true for any cyclic quadrilateral ADBC, not necessarily with perpendicular diagonals $AB \perp CD$.) D is a reflection of the orthocenter H of the triangle $\triangle ABC$ in the line AB, FH = FD. By symmetry, FD' = FD, hence FH = FD = FD' and the triangle $\triangle DD'H$ has right angle $\angle DD'H = 90^\circ$. But D' is a reflection of D in OF, hence $DD' \perp OF$, so that $OF \perp FP$ are midlines of this right angle triangle, i.e., $HD' \perp FP$. Consequently, the quadrilateral FHPD' is a kite, which means that the triangles $\triangle FHP \cong \triangle FD'P$ are (oppositely) congruent and $\angle FHP = \angle FD'P$. But obviously, $\angle FD'P \equiv \angle C'D'B' = \angle CDB = \angle CAB$, which is what we were supposed to prove. Butterfly theorem can be proved in various ways, synthetically or by trigonometry. For example, see http://www.cut-the-knot.org/pythagoras/Butterfly.shtml.

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Unfortunately, there is a trigonometric solution. And I will write it in $\LaTeX$, tenaciously. Let $FP$ intersects the circumcircle of $\triangle AFC$ at $J$. We should show $H,P,J,C$ are concyclic because $\angle FAC = \angle FJC$ and we are asked to show $\angle FHP = \angle FAP$. This yields $FP \cdot FJ = FH \cdot FC = AF \cdot FB$. Let $\angle FCA = \alpha$, $\angle FCB = \beta$, and $\angle AFP = \theta$. $\frac {AF}{FP} = \frac {\sin (90^{\circ} + \alpha - \theta)} {\sin (90^{\circ} - \alpha)}$ $\frac {FC}{FJ} = \frac {\sin (90^{\circ} - \alpha )} {\sin (\alpha+ \theta)}$ $AF \cdot FC = FP \cdot FJ \cdot \frac {\cos(\alpha - \theta)}{\sin(\alpha+\theta)}$ We will show $\frac {FC}{BF} = \frac {\cos(\alpha - \theta)}{\sin(\alpha+\theta)} = \frac {\cos \beta} {\sin \beta}$. Let $R=1$. Thus $AC = 2\cos \beta, BC=2\cos \alpha, BF=2\cos \alpha \sin \beta$, $AF = 2\cos\beta \sin \alpha, AB= 2\sin(\alpha + \beta)$. So $MF = \sin(\beta - \alpha)$ and $OM = \cos (\alpha+\beta)$. And $\angle FOM = \angle AFP = \theta$. Then $\tan \theta = \frac {\sin(\beta - \alpha)}{\cos (\alpha+\beta)}$. $\frac {\cos(\alpha - \theta)}{\sin(\alpha+\theta)} = \frac {\cos \alpha \cos \theta + \sin \alpha \sin \theta}{\sin \alpha \cos \theta + \cos \alpha \sin \theta} = \frac {\cos \alpha + \sin \alpha \tan \theta}{\sin \alpha + \cos \alpha \tan \theta}$. $\Rightarrow \frac {\cos \alpha + \sin \alpha \frac {\sin(\beta - \alpha)}{\cos (\alpha+\beta)}}{\sin \alpha + \cos \alpha \frac {\sin(\beta - \alpha)}{\cos (\alpha+\beta)}} = \frac {\cos \alpha \cos (\alpha + \beta) + \sin \alpha \sin(\beta - \alpha)}{\sin \alpha \cos (\alpha + \beta) + \cos \alpha \sin(\beta - \alpha)}$ $\Rightarrow \frac {\cos \beta (\cos^2 \alpha - \sin^2\alpha)}{\sin \beta (\cos^2 \alpha - \sin^2\alpha)} = \frac {\cos \beta} {\sin \beta}$ $Q.E.D$

Since $\angle HFA=\angle OFP=90^{\circ}$ and $\angle HAF=\angle OAP,$ it follows that $O,H$ are isogonal conjugates with respect to $\triangle APF.$ Consequently, if $M,N$ denote the midpoints of $AB,AC,$ then $\triangle FNM$ is the pedal triangle of $O$ with respect to $\triangle APF$ $\Longrightarrow$ $HP \perp FN$ $\Longrightarrow$ $\angle FHP=\angle NFA.$ Since $\triangle ANF$ is N-isosceles, then $\angle FHP=\angle BAC.$

Luis González wrote: ...$\triangle FNM$ is the pedal triangle of $O$ with respect to $\triangle APF$ $\Longrightarrow$ $HP \perp FN$... why? I think this is not a useful reason to ensure that perpendicularity. The real reason (for me, at least) is: $\measuredangle ONP=\measuredangle OFP=90\Longrightarrow NPFO$ is cyclic, so $\measuredangle FNP=\measuredangle FOP$ but $\measuredangle NPH=\measuredangle OPF$. Finally $\measuredangle FNP+\measuredangle NPH=\measuredangle FOP+\measuredangle OPF=90$, so $HP \perp FN$. (note that in my argument, there was no need to construct either point $M$ nor pedal triangle of $O$.)

Generalization: Triangle $ABC$, and point $F$, such that $\angle BFC=\angle CFA=\gamma$, $\angle FAC=\beta$, point $H$ is on $CF$ and $\angle FHB=\beta$, point $P$ is on $AC$ and $\angle PHF=\beta$, point $O$ with $\angle CBO=\angle OCB= \alpha$, $\alpha+\beta+\gamma=180$. Then prove that $\angle PFA=\angle OFC$.

MY SOLUTION: Let P' be the reflection of P wrt CF then P' is the isogonal conjugate point of O wrt ACF

To avoid using the Butterfly theorem as well as advanced geometry, we can do this way: Let $CF$ cuts the circumcircle at $D$. On $AC$ let’s choose point $P'$ so that $\angle FAP' = \angle BAC$, $P'F$ cuts $BD$ at $Q$ and cuts the circumcirle at $M$ and $N$. Easy to see that $HP'\parallel BD$. As $DF = FH$ is a well known property, $QF = FP'$. If we can prove that $P'M = NQ$ then $F$ is the midpoint of $MN$ and $OF\perp MN$, thus $P'\equiv P$ and we are done. Now using sinus theorem we have $\frac{P'C}{sinF_{1}} = \frac{FP'}{sinC_{1}} $ and $\frac{QD}{sinF_{1}} = \frac{QF}{sinD_{1}}$, thus $\frac{P'C}{QD} = \frac{sinD_{1}}{sinC_{1}}$. Similarly, we get $\frac{BQ}{AP'} = \frac{sinA_{1}}{sinB_{1}}$ Therefore, $\frac{BQ}{AP'} = \frac{P'C}{QD}$, or $BQ.QD = AP'. P'C$ But notice that $QD. BQ = NQ. QM$ and $AP'. P'C = P'M. P'N$, it follows that $NQ = P'M$ $F$ is indeed the midpoint of $MN$ and we are done.

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Let the line through $O$ perpendicular to $AC$ meet $AC,CF,AB$ at $M,L,K$ respectively, thus $\angle OKA=90-\angle A=\angle ACF$ and $M$ is the midpoint of $AC$. From the first result we deduce that $\triangle ACF\sim \triangle LKF$. Since $\angle PFO=\angle CFA=90$, thus $\angle AFP=\angle LFO$, which means that $O,P$ are corresponding points concerning similar triangles $LKF$ and $ACF$. Now let $H'$ be the point that corresponds to $H$, thus $H'$ is on $FK$ and $\triangle FPH\sim \triangle FOH'$ $\Rightarrow$ we now just need to prove $\angle OH'F=\angle A=\angle MFA$ $\iff$ $OH'\parallel MF$ $\iff$ $\frac {MO}{OK}=\frac {FH'}{KH'}=\frac {FH}{CH}(*)$. Since $\triangle AMK\sim \triangle AFC$ and $\angle HAF=\angle MAO$, thus $O,H$ are two corresponding points concerning those two triangles, which means that (*) is true. Q.E.D

Let $DEF$ be the orthic triangle of $\triangle ABC$.Since $\angle (OF, AF) = \angle FPH$ and we already have that $O, H$ as isogonal wrt $\angle FAP$, we get $H, O$ are isogonal conjugates wrt $\triangle AFP$. If we let $M$ be the midpoint of $AC$, then note that $AH \perp FM$ (since they are the feet of the pedals from $O$). Now, $M$ is the centre of $DFAC$ so $\angle MFC = 90 - A$ so $\angle FHP = \angle BAC$ as desired.

@mathreyes It is well-known that for two isogonal conjugates $X, Y$, we have $AX, BX, CX$ is perpendicular to the sides of the pedal triangles. I realise now that my solution is basically the same as Luis's, sorry I posted on impulse hehe

duanby wrote: MY SOLUTION: Let P' be the reflection of P wrt CF then P' is the isogonal conjugate point of O wrt ACF there's a typo, it should be $\triangle BCF$ instead of $\triangle ACF$.

Dear Mathlinkers, see http://www.artofproblemsolving.com/community/c6t48f6h1167200_angle_equal Sincerely Jean-Louis

[asy][asy] size(8cm); pair A=(0,0), B=(120,0), C=(34.5,140.9), F=(34.5,0), O=(60,60), P=(3.3,13.3), Q=(136,97.7), H=(34.5,20.9), X=(19.4,-14.5), Y=(34.4,-20.9); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NW); label("$Q$", Q, NE); label("$H$", H, NW); label("$O$", O, NW); label("$P$", P, NW); label("$H_1$", X, SW); label("$H_2$", Y, S); draw(A--B--C--cycle, linewidth(0.5)+blue); draw(circle(O, 60sqrt(2)), linewidth(0.4)+blue); draw(C--F, linewidth(0.5)+blue); draw(F--P, linewidth(0.5)+blue); draw(X--O--F--cycle, linewidth(0.5)+red); draw(F--Y--O, linewidth(0.5)+red); draw(F--Q--C, linewidth(0.4)+dashed+blue); draw(circle((17.2,2.8), 17.42), linewidth(0.4)+dashed+grey); dot(A); dot(B); dot(C); dot(F); dot(O); dot(P); dot(Q); dot(H); dot(X); dot(Y); [/asy][/asy] Let $H_1$ and $H_2$ be the reflections of $H$ over $\overline{PF}$ and $\overline{AB}$ respectively, and let $\theta=\angle BFO$. Clearly $\angle PFC=\angle PFH_1=\theta$ too. Lemma: $H_1\in (ABC)$ Proof: It is well-known that $H_2\in(ABC)$, so it suffices to prove that $OH_1=OH_2$. Clearly, $H_1F=H_2F$, $OF=OF$, and $$\angle OH_1F=90^\circ+\theta=\angle HFO,$$so $\triangle OFH_1\cong\triangle OFH_2\implies OH_1=OH_2$, as desired. Now, let $Q$ be the second intersection of $\overline{H_1F}$ and $(ABC)$. Since $H_1F=H_2F$, we must have that $CF=QF$ as well. This yields that $$\angle FCQ=\frac{180^\circ-\angle CFQ}{2}=\frac{180^\circ-(180^\circ-2\theta)}{2}=\theta=\angle AFP,$$or that $\overline{CQ}\parallel\overline{PF}$. Finally, this must mean that $$\angle PAH_1=180^\circ-\angle CQF=180^\circ-\angle PFH_1,$$so $PAH_1F$ is cyclic and $$\angle BAC=\angle FH_1P=\angle FHP.$$Yep...

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%200.pdf p. 51... Sincerely Jean-Louis

IMOSL 1996 G3 wrote: Let $O$ be the circumcenter and $H$ the orthocenter of an acute-angled triangle $ABC$ such that $BC>CA$. Let $F$ be the foot of the altitude $CH$ of triangle $ABC$. The perpendicular to the line $OF$ at the point $F$ intersects the line $AC$ at $P$. Prove that $\measuredangle FHP=\measuredangle BAC$. We use phantom points Notations: Let $P'$ be a point on $AC$ such that $\angle FHP'=\angle A$. Now, let $H'$ be the reflection of $H$ on $AB$. Let $\angle AFP'=x\implies \angle P'FH=90-x$. Let $X=H'B\cap P'F$. Two line proof: Clearly, $H'B\| P'H$ and so, $\triangle PHF\equiv \triangle XH'F$ and so, $F$ is the midpoint of $XP'$. Now, obviously sine rule in triangles $XH'F,XFB,AFP',FP'C$ gives power of $X=$ power of $P'$ and so, $XOP'$ is isosceles and thus, $\angle OFP'=90\implies P=P'$. QED

Let $\Gamma$ be the circle with center $O$ of triangle $ABC$, Extend $CF$ intersect $\Gamma$ at $G$, $FP$ intersect $GB$ at $I$, and The point $F$ be a point on chord $DE$ of $\Gamma$ such that $OF \perp DE$. Then We have $F$ is midpoint of $DE$ Since $AB$ intersect $GC$ at F By Butterfly's Theorem, We get $IF=IP$ Since $GBCA$ is cyclic quadrilateral of $\Gamma$ Implies $\angle{GBA}=\angle{GCA}$ $\angle{BGC}=\angle{BAC}$. and $\angle{GCA}=\angle{FCA}=90^\circ -\angle{BAC}=\angle{ABH}=\angle{FBH} =\angle{GBA} = \angle{GBF}$ Since $GH \perp BF$ We deduce $GF=FH$ Then $\triangle{GFI} \cong \triangle{HFP}$ $(S.A.S)$ We obtained $\angle{BAC}=\angle{FGI}=\angle{FHP}$ The result Follows. $\blacksquare$

Reflect $H$ over $AB$ to $H_C$. It is a very well known fact that $H_C$ lies on $(ABC)$. Now let $PF$ meet $BH_C$ at $Q$. By Butterfly theorem, we have $PF$=$FQ$. Now notice that $\triangle PFH \cong \triangle QFH_C$ by SAS congruency. Thus $\angle CAB=\angle CH_CB=\angle FHP$. Our proof is thus complete.

Observe that $\angle BAH = \angle BCH = \angle CAO = \angle BAO = 90^{\circ} - \angle ABC$. Now let $K$ be a point on $AH$ such that $\angle KCH = \angle BCH$. Then $O$ and $K$ are isogonal conjugates in triangle $ACD$, thus $\angle ADO = \angle CDK$. On the other hand, $\angle ADO = 180^{\circ} - \angle EDO - \angle BDE = 90^{\circ} - \angle BDE = \angle CDE$. Hence $\angle CDK = \angle CDE$ and together with $\angle KCD = \angle ECD$ it follows that $\triangle KCD \cong \triangle ECD$ and that $CD$ is the perpendicular bisector of $KE$. But then $\angle DHE = \angle DHK = 90^{\circ} - \angle BAH = \angle ABC$ and we are done.

This is direct use of Butterfly Thorem. Let $H'$ be the reflection of $H$ over $AB$ which we know lies on $(ABC)$. Let the line $FP$ intersect $(ABC)$ at $X, Y$. Because $O \perp FP$ it's clear that $F$ is the midpoint of $XY$. Let $Z$ be the intersection of $H'$ and $B$, then using Butterfly Theorem we know that $FP=FZ$. Now because $H'$ is the reflection of $H$ over $AB$, we know $H'F=FH$ so $H'BHP$ is a parallelogram $\implies H'B \parallel HP \implies \angle BAC = \angle CH'B = \angle FHP$ and we're done. $\square$

Nice bashing here we goooooooooooo: Take $(ABC)$ the unit circle. Then $h=a+b+c$ and $f=\dfrac{ab+bc+c^2-ab}{2c}$ The equation of line $OF$ is $\overline{z}=\dfrac{\overline{f}}{f}\cdot z=\dfrac{ab+bc+ca-c^2}{ab(ac+bc+c^2-ab)}\cdot z$. Consider $X$ and $Y$ the two intersections of line $OF$ with the unit circle. Then $x$ and $y$ are the solutions of the equation $\dfrac{1}{z}=\dfrac{ab+bc+ca-c^2}{ab(ac+bc+c^2-ab)}\cdot z\iff z^2=\dfrac{ab(ac+bc+c^2-ab)}{ab+bc+ca-c^2}$, which by Viete's implies that $x+y=0$ and $x\cdot y=-\dfrac{ab(ac+bc+c^2-ab)}{ab+bc+ca-c^2}$. $P\in AC\iff P$ is its own projection on chord $AC\iff p=a+c-ac\overline{p}$. $OF\perp FP\iff F$ is the projection of $P$ on chord $XY\iff f=\dfrac{1}{2}(p+x+y-xy\overline{p})\iff$ $\iff\dfrac{ac+bc+c^2-ab}{c}=a+c-(ac+xy)\overline{p}\iff \dfrac{bc-ab}{c}=-(ac+xy)\overline{p}\iff$ $\iff \left(ac-\dfrac{ab(ac+bc+c^2-ab)}{ab+bc+ca-c^2}\right)\overline{p}=\dfrac{b(a-c)}{c}\iff \dfrac{a(a-c)(b^2+c^2)}{ab+bc+ca-c^2}\cdot \overline{p}=\dfrac{b(a-c)}{c}\iff$ $\iff \overline{p}=\dfrac{b(ab+bc+ca-c^2)}{ac(b^2+c^2)}\iff p=\dfrac{c(ac+bc+c^2-ab}{b^2+c^2}$. Ok, now we just need to prove the angle condition, which is equivallent to proving that $\widehat{BAC}+\widehat{CHP}=180^\circ\iff \dfrac{b-a}{c-a}\cdot\dfrac{c-h}{p-h}\in \mathbb{R}$ But $c-h=-a-b$ and $p-h=\dfrac{ac^2+bc^2+c^3-abc}{b^2+c^2}-a-b-c=\dfrac{-abc-ab^2-b^2c-b^3}{b^2+c^2}=(-b)\cdot\dfrac{(a+b)(b+c)}{b^2+c^2}$, so $\dfrac{b-a}{c-a}\cdot\dfrac{c-h}{p-h}=\dfrac{b-a}{c-a}\cdot\dfrac{b^2+c^2}{b(b+c)}$, which is clearly real after conjugating, so we're done.