The problem is symmetric for $FD$ and $EG$ and we see, that they meet at $CP$. So, let's prove that the intersection of $CP$ and $FD$ is a point $T$, which lies on $\omega$, the circumcircle of $ABC$. Then by the symmetry we are done. But let's use the phantom point $T'$ which is the intersection of $CP$ and $\omega$. Then $FT'$ intersects $AB$ at $D$. It is enough to prove that $AF||DP$. we know that $\angle{IAC}=\angle{BAI}=\angle{IBA}$, so $\omega_1$, the circumcircle of $ABI$ is tangent to $AC$. Because $T' \in \omega$, we have $\angle{PT'A}=\angle{CT'A}=\angle{CBA}=\angle{BAC}=\angle{PFC}$, so $AT'FP$ is cyclic. Next, $$\angle{PBD}=\angle{PBA}=\angle{PAC}=\angle{PAF}=\angle{PT'F}=\angle{PT'D},$$so $T'DPB$ is cyclic, so we have $\angle{DPT'}=\angle{DBT'}=\angle{ABT'}=\angle{ACT'}$, so $DP$ and $AC (AF)$ are parallel, done.

AlastorMoody wrote:

Where did you use the fact that P lies on circumcircle of triangle AIB?

BOBTHEGR8 wrote: Where did you use the fact that P lies on circumcircle of triangle AIB? The circumcircle of $\odot (AIB)$ is tangent to $AC,BC$ at $A,B$. Hence, $$\angle FAP=\angle PBD$$

Let $CP$ intersect $\odot ABC$ again at $K$. It is enough to show that $K,D,F$ are collinear, as $AFPD$ is a parallelogram it is equivalent to showing that $KF$ bisects $AP$. $\angle CFP=\angle CBA=\angle PKA $ , hence $AFPK$ is cyclic $\implies \angle FAP=\angle FKP$. Let $M$ be mid point of $AP$. $P\in\odot AIB \implies \angle FAP=\angle PBA=\angle PNM$, where $N$ is center of $\odot AIB$. We know that ,as $AC=BC$, $N$ is diametrically opposite point of $C$ in $\odot ABC$. $\therefore\angle NKP=90=\angle NMP$, hence $PMKN$ is cyclic . $\therefore \angle MKP=\angle MNP=\angle FAP=\angle FKP$. Hence F,M,K are collinear. Hence proved.

Trivial problem [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 = -10.892159678721823, xmax = 4.033260360198622, ymin = -3.0135777556749677, ymax = 6.843014568728982; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw((-9.017759525782076,1.807139080272452)--(-6.660735846716839,6.216724024042758), linewidth(1.2)); draw((-6.660735846716839,6.216724024042758)--(-4.154207988979557,1.890372431951916), linewidth(1.2)); draw((-4.154207988979557,1.890372431951916)--(-9.017759525782076,1.807139080272452), linewidth(1.2)); draw(circle((-6.562814535980677,0.4949150171122163), 2.7836463307840944), linewidth(1.2)); draw((-7.267200977051398,3.18796656443237)--(-7.995937463962496,1.8246262343370672), linewidth(1.2)); draw((-7.267200977051398,3.18796656443237)--(-6.492241301586164,1.850360036180239), linewidth(1.2)); draw(circle((-6.611775191348758,3.3558195205774877), 2.8613234217267256), linewidth(1.2) + dtsfsf); draw((-6.660735846716839,6.216724024042758)--(-7.758505431528683,0.7343349259290981), linewidth(1.2)); draw((-8.289023038870976,3.1704794103677547)--(-7.758505431528683,0.7343349259290981), linewidth(1.2)); draw((-7.758505431528683,0.7343349259290981)--(-4.92916766444479,3.2279789602040467), linewidth(1.2)); draw((-8.289023038870976,3.1704794103677547)--(-4.92916766444479,3.2279789602040467), linewidth(1.2)); draw((-7.267200977051398,3.18796656443237)--(-9.017759525782076,1.807139080272452), linewidth(1.2)); draw((-7.267200977051398,3.18796656443237)--(-4.154207988979557,1.890372431951916), linewidth(1.2)); draw(circle((-6.071908797588097,1.6726223472918647), 1.9300237021102353), linewidth(1.2) + linetype("2 2")); draw(circle((-7.7581067154439785,2.010259234252722), 1.2759243706214092), linewidth(1.2) + linetype("2 2")); /* dots and labels */ dot((-9.017759525782076,1.807139080272452),dotstyle); label("$A$", (-8.982868933483298,1.9001806597358508), NE * labelscalefactor); dot((-4.154207988979557,1.890372431951916),dotstyle); label("$B$", (-4.117569674042996,1.9874071404827884), NE * labelscalefactor); dot((-6.660735846716839,6.216724024042758),linewidth(4pt) + dotstyle); label("$C$", (-6.618062122121876,6.290580190665044), NE * labelscalefactor); dot((-6.61044604370653,3.2781538021178283),linewidth(4pt) + dotstyle); label("$I$", (-6.569602966151356,3.3539553388514776), NE * labelscalefactor); dot((-7.267200977051398,3.18796656443237),dotstyle); label("$P$", (-7.2286474873504405,3.2861125204927486), NE * labelscalefactor); dot((-7.995937463962496,1.8246262343370672),linewidth(4pt) + dotstyle); label("$D$", (-7.955534826908255,1.9001806597358508), NE * labelscalefactor); dot((-6.492241301586164,1.850360036180239),linewidth(4pt) + dotstyle); label("$E$", (-6.453300991822106,1.9292561533181634), NE * labelscalefactor); dot((-8.289023038870976,3.1704794103677547),linewidth(4pt) + dotstyle); label("$F$", (-8.24628976273138,3.2473451957163317), NE * labelscalefactor); dot((-4.92916766444479,3.2279789602040467),linewidth(4pt) + dotstyle); label("$G$", (-4.892916169571332,3.305496182880957), NE * labelscalefactor); dot((-7.758505431528683,0.7343349259290981),linewidth(4pt) + dotstyle); label("$H$", (-7.722930878249754,0.8146955659961829), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that we have $\Delta CFG$ homothetic to $\Delta PDE \implies PD=PE=GB=FA \implies PGBD$ is an iscoseles trapezoid. Similiarly $AFPE$ is iscoseles trapezoid. Now let $CP \cap \odot(ABC)=H$. Note that $\angle CHA=\angle CBA=\angle PEA\implies H \in \odot(AFPE)$. Similiarly $H \in \odot PGBD \implies \angle CAP=\angle PBA\equiv \angle PBD=\angle PHD$ But since $\angle PAC=\angle PHF \implies \{D,F,H\}$ is a collinear triplet. Similiarly $\{E,G,H\}$ is a collinear triplet which implies the result. $\blacksquare$

The intersection point is actually the homothetic center of triangles \(CFG\) and \(PDE\) and is also the \(P\)-Dumpty point in \(APB\).

here is the older post back from 2004

Here's an alternative approach using 2019 AIME I P15. I've drawn $P$ outside $\triangle ABC$ to make the connection more explicit; the proof is the same. Reinterpret the problem with reference triangle $PAB$; then $C$ is the interesection of the tangents to $(PAB)$ at $A$ and $B$. Let line $PC$ intersect $(ABC)$ at $T$, $\overline{AB}$ at $K$, and $(PAB)$ at $Q$. We will show that $T$ is the desired intersection point. [asy][asy] size(300); defaultpen(fontsize(10pt)); pair A, B, C, O, I, P, D, E, F, G, T, Q, K; O = (0,0); P = dir(120); A = dir(215); B = dir(325); C = extension(A, rotate(90, A)*O, B, rotate(90, B)*O); I = incenter(A, B, C); D = extension(A, B, P, P+C-A); E = extension(A, B, P, P+C-B); F = extension(C, A, P, P+A-B); G = extension(C, B, P, P+A-B); T = extension(D, F, E, G); Q = IP(P--C, circumcircle(P, A, B), 1); K = extension(P, Q, A, B); draw(A--B--P--cycle, orange); draw(C--F--G--cycle, orange); draw(A--E--P--D, orange); draw(circumcircle(A, B, P), red); draw(circumcircle(A, B, C), lightblue); draw(D--F^^E--G, heavygreen+dashed); draw(C--P, heavycyan+dotted); dot("$A$", A, dir(220)); dot("$B$", B, dir(0)); dot("$C$", C, dir(270)); dot("$I$", I, dir(270)); dot("$P$", P, dir(90)); dot("$D$", D, dir(270)); dot("$E$", E, dir(180)); dot("$F$", F, dir(130)); dot("$G$", G, dir(50)); dot("$T$", T, dir(70)); dot("$Q$", Q, dir(225)); dot("$K$", K, dir(240)); [/asy][/asy] Claim: We have $CQ/CT = CK/CP$. Proof. Since $T$ is the midpoint of the $P$-symmedian chord in $\triangle PAB$, the AIME problem implies $(TAK)$ and $(TBK)$ tangent to $(PAB)$. In particular, we have $$CP\cdot CQ = CA^2 = CK\cdot CT \implies \frac{CQ}{CT} = \frac{CK}{CP},$$as desired. $\blacksquare$ Observe that $\triangle PDE$ and $\triangle CFG$ are homothetic with center $S = \overline{CP}\cap \overline{DF}\cap\overline{EG}$, and that $PS/SC = DE/FG$. However, since $T$ is the midpoint of $\overline{PQ}$, we also have $$\frac{PT}{TC} = 1 - \frac{CQ}{CT} = 1 - \frac{CK}{CP} = \frac{KP}{CP} = \frac{DE}{FG}.$$(For the last equality, we used $\triangle PDE\cup K \sim \triangle CFG\cup P$.) So $S = T$, which completes the proof.

Here is my solution using barycentric coordinates Let $ABC$ be the reference triangle with the usual coordinates for each vertex and $AB=c$ , $BC=AC=a$ Let $P=(i:j:k)$ with $i+j+k=1$ subbing the coordinates of each point , $A$ ,$B$ and $I$ we get that the equation of the circumcircle of $AIB$ is $$-a^2zy-a^2xz-c^2xy+a^2z(x+y+z)=0$$so we have $$-a^2jk-a^2ik-c^2ij+a^2k=0 \iff a^2k(1-j-i)-c^2ij=0 \iff a^2k^2-c^2ij=0 ~~~(1)$$ The point in infinity that lies in $AB$ is $(1:-1:0)$, in $AC$ is $(1:0:-1)$ and in $BC$ is $(0:1:-1)$ so we compute the coordinates of $D,E,F$ and $G$ and we find $ D=(i+k:j:0)$ , $E=(i:j+k:0)$, $F=(i+j:0:k)$ and $G=(0:i+j:k)$ Let $X = (DF) \cap (EG)$ and $X=(1:e:f)$ using colinearity criteria, we see that $X$ must fulfills the equations : $$-(i+j)fj+k(j-e(i+k))=0$$and $$(i+j)fi+k(k+j-ei)=0$$solving these equations , we find $e=\frac{j}i$ and $f=\frac{-k^2}{i(i+j)}$ so $X=(i(i+j):j(i+j):-k^2)$ so by subbing in the equation of the circumcircle of $ABC$ , we find $$a^2j(i+j)k^2+a^2i(i+j)k^2-c^2ij(i+j)^2=(i+j)(a^2k^2(i+j)-c^2ij(i+j))=(i+j)^2(a^2k^2-c^2ij)=0$$the last equality is because of $(1)$ so $X$ lies on the circumcirle of $ABC$

Does there exist a solution whith moving points?

Note that $(IAB)$ is tangent to $AC,AB$, so $\angle APD=\angle PAF=\angle PBA$, and similarly $\angle EPB=\angle DAP$. So, $DAFP\sim EPGB$. However, as $DE\parallel FG$, the center of the spiral similarity is at $Q=FD\cap GE$. Now, by the spiral, $\angle BAQ=\angle QPE$ and $\angle EBQ=\angle QPE$. So, $\angle AQB=180-(\angle DPQ+\angle QPE)=180-\angle C$, and $Q$ lies on $(ABC)$, as desired.

Here's a solution I don't think is here. Despite remembering a hint from Mathematical Olympiad Challenges from when I first saw this problem back in 2012, I'm happy I finally solved it Also first G5? Let $T$ and $Q$ be the intersection points of $P$ with $\odot(ABC)$ and $\odot(AIB)$, respectively. We claim $T$ is the desired intersection point. [asy][asy] size(300); defaultpen(linewidth(0.75)+fontsize(10)); pair C = dir(90), A = dir(215), B = dir(325), T = dir(295), I = incenter(A,B,C), P = intersectionpoint(C--T,circumcircle(A,I,B)), Q = 2*T-P; pair R = (-10,P.y), S = (10,P.y), F = intersectionpoint(R--S,A--C), G = intersectionpoint(R--S,B--C); pair D = intersectionpoint(F--T,A--B), E = intersectionpoint(G--T,A--B); draw(A--B--C--A,orange); draw(circumcircle(A,B,C),red); draw(circumcircle(D,T,B),red+linetype("3 3")); draw(F--G^^D--P--E,lightblue); draw(circumcircle(A,P,B),heavygreen); draw(C--Q,purple); draw(F--T^^A--Q,brown+linetype("3 3")); dot("$A$",A,SW,linewidth(3.3)); dot("$B$",B,SE,linewidth(3.3)); dot("$C$",C,N,linewidth(3.3)); dot("$P$",P,NE,linewidth(3.3)); dot("$I$",I,N,linewidth(3.3)); dot("$F$",F,W,linewidth(3.3)); dot("$G$",G,dir(0),linewidth(3.3)); dot("$D$",D,SW,linewidth(3.3)); dot("$E$",E,dir(270),linewidth(3.3)); dot("$Q$",Q,dir(C--Q),linewidth(3.3)); dot("$T$",T,SE,linewidth(3.3)); [/asy][/asy] Observe that $\angle PDE = \angle CAB = \angle CTB$, so quadrilateral $PDTB$ is cyclic. In turn, $\angle PTD = \angle PBD = \angle PQA$, so $DT\parallel AQ$. Now recall by Fact 5 that the center of the circle $\odot(AIB)$ is the midpoint $M$ of the arc $\widehat{AB}$ of $\odot(ABC)$. Since $\angle CTM = 90^\circ$, $T$ is the midpoint of $\overline{PQ}$. Thus $TD$ also passes through the midpoint of $\overline{AP}$. But line $FD$, being a diagonal of parallelogram $AFPD$, does so too; in turn, $F$, $D$, and $T$ are collinear. Analogously, $G$, $E$, and $T$ are collinear, so $FD$ and $GE$ do, indeed, meet at $T$.

Is this some sort of joke? This is much easier than G1. Let $FD$ and $GE$ intersect at $K.$ The crucical claim is that $(AFPEK)$ and similarly $(BGPDK)$ are cyclic. Note that $P$ lies on $(AFE)$ since $\angle PFA=180^{\circ}-\angle CFG=180^{\circ}-\angle PEA.$ Now we claim that that $\triangle APD \sim \triangle PEB.$ Note that $\angle ADP=\angle PEB$ and $\angle APD+\angle PAD=\angle ABC$ while $\angle APD+\angle BPE=\angle ABC$ as well, so $\angle BPE=\angle DAP.$ This implies that $\triangle PDF\sim \triangle PGE,$ or that $\angle PFD+\angle PGE=\angle ACB+\angle ABC,$ or $\angle FKG=\angle ABC.$ Since $\angle AFK=\angle ABC=\angle FKG,$ $K$ lies on $(AFPE).$ Now note that $\angle AKB=\angle AKF+\angle BKG+\angle DKE=\angle APF+\angle BPG+\angle ACB=\angle ABC+\angle ACB,$ so $K$ lies on $(ABC).$

If only recent G5's were this easy... Let $Z = CP \cap (ABC)$. Since $\angle PGC = \angle ABC = \angle AZC$, we have $PGBZ$ cyclic and similarly $FPZA$ is cyclic as well. Define $E' = GZ \cap AB$. Then, we have $\angle E'PB = \angle APB - \angle APE' = 90 + \frac{\angle ACB}{2} - \angle APE' = 90 + \frac{\angle ACB}{2} + \angle AZE' - 180 = \angle AZE' - \angle ABC = \angle AZE - \angle AZC = \angle PZG = \angle PBG$ and so $PE'$ is parallel to $GB$, which means $E' = E$ So, $DF$ and $EG$ intersect at $Z$, which is indeed on the circumcircle of $ABC$, as desired. $\blacksquare$

Let $CP$ meet $(AIB)$ at $X$ and $(ABC)$ at $Y$, now it's enough to prove $\overline{Y-E-G}$ and $\overline{Y-D-F}$ are collinear. Claim: $YDPGB$ and $YEPFA$ are cyclic. Proof: $$\angle BYP = \angle BYC = \angle BAC = \angle BAF = \angle BDP$$and since $PDBG$ is an isosceles trapezium hence $YDPGB$ is cyclic. In a similar manner, it's true for $YEPFA$. $$\angle PYG = \angle PBG = \angle PXB$$Hence $YG \parallel XB$. $$\angle PYE = \angle PAE = \angle PAB = \angle PXB$$Hence $YE \parallel XB$. Thus $\overline{Y-E-G}$ are collinear. Similarly $\overline{Y-D-F}$ are also collinear, hence $FD \cap GE = Y \in (ABC)$.

Let $CP \cap (AIB)=J$ and $(PFA) \cap (PGB)=H$. Claim 1: $H$ lies on $(ABC)$ Proof: By angle chasing $\angle AHB=\angle AHP+\angle PHB=\angle CFP+\angle CGP=180-\angle ACB$ thus $ACBH$ is cyclic. Claim 2: $A,P,H,J$ are colinear. Proof: First by the parallelograms & homothety formed on the construction $AF=DP=PE=GB$ thus $AFGB$ is a isosceles trapezoid and now by radical axis on $(PFA),(PGB),(AFGB)$ we have the desired colinearity. Claim 3: $G,E,H$ and $F,D,H$ are colinear. Proof: By angle chasing $\angle PJB=\angle PBG=\angle PHG$ thus $GH \parallel BJ$ and doing the same thing on the $A$ side we have that $AJ \parallel FH$ and now by homothety its done!. By Claim 3 we solved the problem

Anger. Note that $DE||FG,CF||DP,CG||PE$ so $\triangle CFG$ is a homothety of $\triangle PDE.$ Thus, $FD,GE,CP$ concur, at say $H.$ By the incenter-excenter lemma, $AC$ is tangent to $(AIB)$ so $\angle FAP=\angle ABP=\angle GPB.$ This implies that $\triangle FAP\sim\triangle GPB.$ Thus, the parallelograms $AFPD$ and $PGBE$ are similar. We can say $\angle AFD=\angle PGE=\angle DEH$ so $AFEH$ is cyclic. Also, $AFPE$ is isosceles trapezoid so $A,F,P,E,H$ are concyclic. Thus, $\angle AHP=\angle CFG=\angle CBA$ so $H$ lies on $(ABC)$ as desired.

First, let $FD$ and $GE$ meet at $M$. Then note that the homothety center of triangles $CFG$ and $PDE$ is $M$, as that is where $GI$ and $FD$ meet. Therefore, $AP$ passes through $M$. Let $AP$ meet $\Gamma$ at $M'$. We will prove that $M' = M$. Extend $CM$ to meet $(AIB)$ at $N$. Let $\angle CBA = a$, and note that $\angle BM'C = a$, and $\angle FGB = 180 - A$, so $M'PGB$ is cyclic. Then, since it is well-known that the circumcentre of $(AIB)$ is the arc midpoint of $AB$ ($O$, say), note that $\angle CBO = a + \angle ACO$, but since it is well-known that $C,I,O$ are collinear, $\angle CBO = 90$. Therefore, $CB$ is tangent to $(AIB)$, so by the alternate segment theorem if $\angle CBP = \theta$ then $\angle BNP = \theta = \angle GM'P$, so $BN || GM'$. Likewise, $AN || FM'$. This is very useful, as now we have that $\triangle BNA \sim \triangle GM'F$, so $\angle GM'F = 180 - (180 - 2\frac{a}{2}) = a$. Now, if we can prove that $\angle GMF = a$, then we will be done as $M'$ is on the same side of $FG$ as $M$ as $FG$ is above $AB$ which is above $M$ and the point where $AM$ meets $\Gamma$, implying that since as $M$ moves on $AP$ the angle changes, there is only one point on a specific side of $FG$ with $\angle GMF = a$, so we would have $M = M'$.We will prove this with congruent triangles. Let $\angle FGE = x$. Then we only need that $\angle DFP = 180 - a - x$. Note that since $DAFP$ and $BIPG$ are both parallelograms, $\angle BIG = x$, $\angle FDA = \angle DFP$. Then, if we can prove that $\angle FDA = 180 - a - x$, we will be done. However, note that if this is true then $\triangle FDA \sim \triangle IGB$, which implies that $\frac{FA}{IB} = \frac{AD}{GB}$. Since $FA = GB$ as $FG || AB$,we need that $(GB)^2 = (GP)(FP)$. However, if we let $Z$ be the intersection of $GP$ with $(AIB)$, then we have that $(GB)^2 = (GP)(GZ)$, so all we need that that $GZ = FP \implies FZ = GP$. However, since $AB || ZP$, since if a trapezoid is cyclic it is isosceles, $AZ = PB$. Also, $\angle BZA = \angle PBA$, so $\angle ZAF = \angle GBP$. Therefore, by $SAS$ congruence, $FZ = PG$, so we are done.

Let $X=\overline{CP}\cap(ABC).$ Notice $AEPF$ is cyclic as $\overline{AE}\parallel\overline{FP}$ and $FA=GB=PE.$ Since $\angle XPE=\angle PCB=\angle XAB,$ $XEPFA$ is cyclic. Similarly, $BGPDX$ is cyclic. We see $\overline{AC}$ is tangent to $(AIB)$ at $A,$ so $$\angle PXD=\angle PBD=\angle PBA=\angle PAF=\angle PXF$$and $X$ lies on $\overline{DF}$ and similarly $\overline{EG}.$ $\square$

Observe that $AFPE,BGPD$ are isosceles trapezoids and by Miquel theorem in $\triangle CFG$ their circumcircles meet at $Q\in \odot (ABC).$ $ABGF$ is cyclic, so by radical axis $C\in PQ.$ But clearly $\odot (APB)$ is tangent to $AC$ so $$\measuredangle PQD=\measuredangle PBD=\measuredangle PAF=\measuredangle PQF\implies Q\in DF.$$Analogously $Q\in EG$ and so we are done.

I'm not sure why people are talking about this being misplaced at g5. To me it was very well suited, pretty difficult even with phantom point, and much much harder than g1-4, those of which were instasolves. Even though the solutions are short it is still difficult to find. First note the cyclicislscelestrapezoids AFPE and BDPG. Let CP intersect the circumcircle of ABC in J. Now note JPE=JCB=JAB, hence JAFPE is cyclic, and by analogous reasoning so is JDPGB. It's easy to see that AC is a tangent (one way by angle chasing, we have CKB+KCB=CAB+1/2C=90 where K is center and also midpoint of arc AB), hence PJD=PBA=CBA-CBP=BAC-BAP=PAC=PJF, hence J lies on DF and analogously EG, as desired. $\blacksquare$

Tried to solve it with some friends. But I eventually deviated from what they got... So here is something which I got! Solution: Let $X = \overrightarrow{CP} \cap \odot(CAB)$. We claim that for any fixed $P$, the desired circles will intersect at $X$. Also define $P' \coloneqq CP \cap \odot(AIB) \ne P$. Re-define $D = FX \cap AB$ and $E = GX \cap AB$. Now all we have to show is that $PD \parallel CA$ and $PE \parallel CB$. [asy][asy] import olympiad; import geometry; size(12cm); defaultpen(fontsize(13pt)); pair A = (-2,0); pair B = (2,0); pair C = (0,5); pair I = incenter(A,B,C); pair P = (-0.6, 1.26);// yeah, its random pair X = intersectionpoints(line(C,P), circumcircle(A,B,C))[1]; pair P_ = 2*X-P; pair F = extension(P,P+A-B, A,C); pair G = extension(P,P+A-B, C,B); pair D = extension(F,X,A,B); pair E = extension(G,X,A,B); draw(A--B--C--A, purple); draw(F--G, purple); draw(C--P_, purple); draw(F--X, purple); draw(G--X, purple); draw(A--P_, purple); draw(B--P_, purple); draw(P--A, purple); draw(P--B, purple); draw(P--D, purple); draw(P--E, purple); draw(circumcircle(A,B,C), red); draw(circumcircle(I,A,B), red); draw(circumcircle(F,X,A), dashed + magenta); draw(circumcircle(D,X,P), dashed + magenta); draw(circumcircle(F,G,X), fuchsia); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(I)); dot("$P$", P, dir(P)); dot("$X$", X, dir(X)); dot("$P'$", P_, SW); dot("$F$", F, dir(F)); dot("$G$", G, dir(G)); dot("$D$", D, dir(210)); dot("$E$", E, dir(320)); [/asy][/asy] Since $\triangle CAB$ is isosceles, we already have that $AC$ is tangent to $\odot(AIB)$. Also $XPFA$ is cyclic since \[\measuredangle AXP = \measuredangle AXC = \measuredangle ABC = \measuredangle PFC.\]Cyclicity of $XBGP$ follows similarly. We now show the following crucial claim. Claim: $\triangle XFG \sim \triangle P'AG$ which in turns give us that $XP$ is symmedian of $\triangle XFG$. To extend even more, $CG$ and $CF$ are tangents to $\odot(XFG)$. Proof: Note that since $(P',P;A,B) = -1$, showing the similarity of triangles is enough. Observe that \[\measuredangle AP'P = \measuredangle PBA = \measuredangle PAC = \measuredangle PXA\]which shows $FX \parallel AP'$. Similarly, we have $XG \parallel P'B$ which proves the similarity. Also since $A$ is the intersection of the perpendicular bisector of $FG$ and $X-$symmedian of $\triangle XFG$, we get that $FC$ and $GC$ are indeed tangents to $\odot(XFG)$. $\square$ Finally we show that $E \in \odot(XPFA)$ and $D \in \odot(XPGB)$. We only show one part as other one follows by symmetry. This is an easy angle chase. \[\measuredangle EAF = \measuredangle GFC = \measuredangle GXF = \measuredangle EXF\]This proves $E \in \odot(AFX)$ as desired. The solution is almost over. Recall $FG \parallel AB$, so we have that $FPAE$ and $GPDB$ are isosceles cyclic trapezoids. By symmetry, we're done! $\blacksquare$

I solved this last summer but feel like posting a write-up right now. See the diagram from post #26 (right above this one). Let the midpoint of arc $AB$ not containing $C$ be $M$, $K = AB \cap CP$, $X = DF \cap EG$, and $P_1$ denote the second intersection of $\overline{CPK}$ and $(AIB)$. We know $M$ is the center of $(AIB)$ by the Incenter-Excenter Lemma. But $CM$ is clearly a diameter of $(ABC)$, so Thales' implies $AB$ is the polar of $C$ wrt $(AIB)$, which means $-1 = (C, K; P, P_1)$. It's easy to see $ABC$ and $DEP$ are homothetic with center $K$, while $XDE$ and $XFG$ are homothetic with center $X$. However, $DPE$ and $FCG$ are clearly homothetic as well, so $XDPE$ and $XFCG$ are homothetic at $X$, giving $X \in \overline{CPK}$. Thus, $XDPEK$ and $XFCGP$ are homothetic figures at $X$, yielding $\frac{XK}{XP} = \frac{XP}{XC}$ or $XP^2 = XC \cdot XK$. Now, the uniqueness of harmonic conjugates implies $P_1$ is the reflection of $P$ over $X$. Thus, $MX \perp \overline{CPP_1}$ follows from $MP = MP_1$, which finishes via Thales'. $\blacksquare$

Let $CP$ intersect (ABC) again at $P'.$ By trivial angle chasing we get $ \angle FAP = \angle DBP$ Observe $AFPEP' $ and $PGBP'D$ are cyclic. By angle chasing observe $ P',D,F$ and $P',E,G$ are collinear. We are done

Here is my solution: Note that $AFPE, PGBD$ are isosceles trapezoids. Let $M$ denote the midpoint of arc $AB$ (not containing $C$). Note that, by Incenter-Excenter Lemma, $M$ is the center of $(AIB)$. Since $CA=CB$, $(AIB)$ is tangent to $CA, CB$ at $A, B$ respectively. Let $X = CP \cap (ABC)$. Then, $$\angle AKP = \angle AKC = \angle ABC = \angle PEA$$Thus, $AFPEK$ is cyclic. Similarly, $DPGBK$ is cyclic. Thus; $$\angle FKP = \angle FAP = \angle PBA = \angle PBD = \angle DKP$$Hence, $F, D, K$ are collinear. Similarly, $G, E, K$ are collinear and we conclude.

bu h did i overcomp,icate. for some reason i saw cyclic quads and couldn't doa nything but i think this sol is nicer ha Let $Q=DF\cap EG$ and let $X$ be the exsimilicenter of $DE$ and $AB$. Let $M$ be the midpoint of arc $AB$. From Monge on $DE$, $FG$, and $AB$ we find that $Q$, $X$, and $C$ are collinear. However $X$ is also the center of homothety sending $\triangle PDE$ to $\triangle CAB$. Thus $X$, $P$, and $C$ are collinear. Thus $Q$ lies on $CP$. Let $P'=CP\cap (AIB)$. Notice that $C$ and $X$ are harmonic conjugates with respect to $P'P$ (here we use the fact that $CA$ and $CB$ are tangent to $(AIB)$). It suffices to show that $Q$ is the midpoint of $P'P$; this will imply $MQ\perp CQ$ or equivalently $Q\in (ABC)$. By a well-known property it suffices to show $QX\cdot QC=QP^2$. Notice that \[\frac{QX}{QP}=\frac{DX}{FP}=\frac{DX}{AD}=\frac{XP}{PC}\implies QX\cdot PC=QP\cdot XP.\]This is equivalent to \[QX\cdot (QC-QP)=QP\cdot (QP-QX)\implies QX\cdot QC=QP^2\]and we are done. $\blacksquare$ wait bruh you use the other "phantom point" definition and angle chase im done i actually found this ,,,, lmao oops well at least the problem is solved.