By all means, this is far from being a "pretty easy" geometry problem. Either that, or I was very dumb last year at IMO when I thought about it. The discussion was about whether choosing A4 (which was actually problem 5 in the contest) or G5 (which is this problem). You all know the result PS this is Hojoo's proposal, too.

Well what I did was to remove I, backward constructed the question, used a bit of power of a point and it came out in a few lines of length calculations, so it didn't seem that hard to me.

Well, here's what I did: By a very quick and easy angle chase I showed that $\angle FXG=\angle B (=\angle A)$, where $X=FD\cap GE$. Since $\angle AGF=\angle AFG=\angle FXG$, we find circle $(FGX)$ to be tangent to $CA$ and $CB$. It's easy to show that circle $(AIB)$ is also tangent to $CA,\ CB$. Let $X=AX\cap (ABC),\ X'=AX\cap (AIB),\ T=CX\cap AB$. Then $\frac{CX'}{CX}=\frac{CB}{CG}$, so all we need in order to show that $X=X$ is to show that $\frac{CX'}{CX}=\frac{CB}{CG}$. This is easy because $CB^2=CX'\cdot CP=CX\cdot CT\Rightarrow \frac{CX'}{CX}=\frac{CT}{CP}=\frac{CB}{CG}$ because $PG||TB$. Here I used the fact that the inversion of pole $C$ and power $CB^2$ turns the circle $(ABC)$ into the line $AB$ and it invariates the circle $(AIB)$, because it's tangent to $CA,CB$ in the invariant points $A,B$. I think we're done.

Here is the solution I found on the exam: Since PE || CB, we have < FPE = < FGB = 180 - < FGC, and since PG || AB, we have < FGC = < ABC, so that < FPE = 180 - < ABC. Since AC = BC, we have < ABC = < BAC, and < FPE = 180 - < BAC = 180 - < FAE. Thus, the quadrilateral FPEA is cyclic, i. e. the points F, P, E and A lie on one circle. Similarly, the points G, P, D and B lie on one circle. Let S be the point of intersection of these two circles (distinct from P). Since the points P, S, A and E lie on one circle, < PSE = < PAE = < PAB = 180 - < APB - < PBA. Since the points A, P, B and I lie on one circle, < APB = < AIB, and 180 - < APB = 180 - < AIB = < IAB + < IBA = < BAC / 2 + < ABC / 2 = < ABC / 2 + < ABC / 2 = < ABC, so that < PSE = 180 - < APB - < PBA = < ABC - < PBA = < PBG. Finally, the concyclic points P, B, G and S yield < PBG = < PSG. Hence, < PSE = < PSG, and the point S lies on the line EG. Similarly, the point S lies on the line DF. Hence, the point S is the point of intersection of the lines DF and EG. Now, we have to show that S lies on the circumcircle of triangle ABC. Well, the cyclic quadrilaterals FPSA and GPSB show < ASP = 180 - < AFP and < PSB = 180 - < PGB, yielding < ASB = < ASP + < PSB = (180 - < AFP) + (180 - < PGB) = < CFG + < FGC = 180 - < FCG = 180 - < ACB, so that the quadrilateral ACBS is cyclic, and our point S does lie on the circumcircle of triangle ABC. Proof complete. I am not a genius and I needed more than an hour to come up with this. Darij

This is my solution. Let circumcircle of $\triangle ABC$ : $w$. $CP \cap w=T$, $TF\cap AB= D'$, $TG \cap AB = E'$. $CT\cap AB= X$ cuz $\angle PTA= \angle ABC= \angle CFG$, $P, F, A, T$ is cyclic. similarly, $P, G, B, T$ is cyclic. Here, $\angle FTP = \angle FAP = \angle PBA$. so $T, B, P, D'$ is cyclic. $\Rightarrow \triangle PXB \sim \triangle D'XT$. cuz $A, T, B, C$ is also cyclic. $\triangle CXB \sim \triangle AXT$ . therefore, $CP:PX = AD' : D'X$. Here, We get $AC \parallel PD'$. This means that $D'= D$. Similarly, $E'=E$. $Q.E.D.$

Let $ABC$ be an isosceles triangle with $AC=BC$, whose incentre is $I$. Let $P$ be a point on the circumcircle of the triangle $AIB$ lying inside the triangle $ABC$. The lines through $P$ parallel to $CA$ and $CB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $AB$ meets $CA$ and $CB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC\ .$ Commentary. The enunciation of this problem conceals vainly in its debut that the circumcircle of the triangle $AIB$ is in fact the circle with the diameter $II_{c}\ ,$ where the point $I_{c}$ is the $C$- exincenter of the triangle $ABC\ .$ Proof. Denote : the circumcircle $w$ of the triangle $ABC$ ; the circle $\delta$ with the diameter $[II_{c}]$ ; the circumcircles $w_{a}$ , $w_{b}$ of the isosceles trapezoids $AFPL$ , $BGPD$ respectively ; the second intersection point $R$ between the line $\overline{FPG}$ and the circle $\delta$ ; the intersection $L\in FD\cap GE\ .$ Prove easily that the lines $CA$ , $CB$ are the tangents from the point $C$ to the circle $\delta\ .$ From the relation $FA^{2}=FR\cdot FP$ obtain $\frac{FA}{AD}=\frac{GP}{PE}\ ,$ i.e. the quadrilaterals $FADP$ , $GPEB$ are similarly as $FADP\sim GPEB\ .$ Thus, $\begin{array}{cc}1\blacktriangleright & \begin{array}{c}\widehat{AEL}\equiv\widehat{GEB}\equiv\widehat{FDP}\equiv\widehat{AFD}\equiv\widehat{AFL}\Longrightarrow\widehat{AEL}\equiv\widehat{AFL}\Longrightarrow L\in w_{a}\\\\ \widehat{BDL}\equiv\widehat{FDA}\equiv\widehat{GEP}\equiv\widehat{BGE}\equiv\widehat{BGL}\Longrightarrow\widehat{BDL}\equiv\widehat{BGL}\Longrightarrow L\in w_{b}\end{array}\Longrightarrow\boxed{\ \widehat{DLE}\equiv\widehat{ABC}\ }\\\\ 2\blacktriangleright & \begin{array}{c}\widehat{ALD}\equiv\widehat{ALF}\equiv\widehat{APF}\\\\ \widehat{BLE}\equiv\widehat{BLG}\equiv\widehat{BPG}\end{array}\Longrightarrow \widehat{ALD}+\widehat{BLE}\equiv\widehat{APF}+\widehat{BPG}\equiv\widehat{AI_{a}B}\equiv\widehat{ABC}\Longrightarrow \boxed{\ \widehat{ALD}+\widehat{BLE}\equiv\widehat{ABC}\ }\end{array}$ $\Longrightarrow$ $\widehat{ALB}\equiv\widehat{ALD}+\widehat{BLE}+\widehat{DLE}=$ $2\cdot\widehat{ABC}=$ $180^{\circ}-\widehat{ACB}$ $\Longrightarrow$ $\boxed{\ \widehat{ALB}+\widehat{ACB}=180^{\circ}\ }$ $\Longrightarrow$ $L\in w\ .$

Ok, here's another solution:

I'll post my solution too . Denote $ \angle CAB = \angle ABC = \alpha$ and $ FD \cap GE = S$. First note that triangles $ \triangle APF$ and $ \triangle PBG$ are similar. This follows from $ \angle PFA = \angle BPA = \angle BGP = 180 - \alpha$. Moreover, $ ADPF \sim PEBG$. Hence $ \angle(FD, GE) = \angle(FP, GB) = \alpha$. Therefore, quadrilaterals $ ASEF$ and $ DSBG$ are cyclic. Hence, $ ASEPF$ and $ DSBGP$ are cyclic too. Thus we have to show that circles $ (ABC)$, $ (APF)$, $ (GPB)$ have a point in common. Therefore, consider $ (ABC) \cap (APF) = T$. A simple angle chase shows that $ \angle ATB = 2 \alpha$ and $ \angle ATB = \alpha$. Hence, quadrilateral $ PTBG$ is cyclic too. The conclusion follows.

darij grinberg wrote: Here is the solution I found on the exam: Since PE || CB, we have < FPE = < FGB = 180 - < FGC, and since PG || AB, we have < FGC = < ABC, so that < FPE = 180 - < ABC. Since AC = BC, we have < ABC = < BAC, and < FPE = 180 - < BAC = 180 - < FAE. Thus, the quadrilateral FPEA is cyclic, i. e. the points F, P, E and A lie on one circle. Similarly, the points G, P, D and B lie on one circle. Let S be the point of intersection of these two circles (distinct from P). Since the points P, S, A and E lie on one circle, < PSE = < PAE = < PAB = 180 - < APB - < PBA. Since the points A, P, B and I lie on one circle, < APB = < AIB, and 180 - < APB = 180 - < AIB = < IAB + < IBA = < BAC / 2 + < ABC / 2 = < ABC / 2 + < ABC / 2 = < ABC, so that < PSE = 180 - < APB - < PBA = < ABC - < PBA = < PBG. Finally, the concyclic points P, B, G and S yield < PBG = < PSG. Hence, < PSE = < PSG, and the point S lies on the line EG. Similarly, the point S lies on the line DF. Hence, the point S is the point of intersection of the lines DF and EG. Now, we have to show that S lies on the circumcircle of triangle ABC. Well, the cyclic quadrilaterals FPSA and GPSB show < ASP = 180 - < AFP and < PSB = 180 - < PGB, yielding < ASB = < ASP + < PSB = (180 - < AFP) + (180 - < PGB) = < CFG + < FGC = 180 - < FCG = 180 - < ACB, so that the quadrilateral ACBS is cyclic, and our point S does lie on the circumcircle of triangle ABC. Proof complete. I am not a genius and I needed more than an hour to come up with this. Darij Sorry to bring a very old topic back but you are really a genius Darij! This solution is so simple and genial!

K=w∩CP,X=KD∩AP.∠PAD=∠PBC obviously.Let ∠BAC=∠ABC=2a,then ∠IAB=∠IBA=a and it follows that ∠AIB=180-2a=∠APB⟹∠PAB+∠PBA=2a=∠BAP+∠PAC⟹∠PBA=∠PAC=∠APD.We know that ∠CAB=∠PDE=∠PDB=∠PKB, and it follows that quadrilateral PDKB is cyclic,so ∠PBD=∠PKD.So, we have AX2=XD•XK=PX2⇒AX=PX,X is the midpoint of AP.We know that FD intersect AP on its midpoint,so F,X,D,K are collinear points.Analogously, K,E and G are collinear,as desired.

Let $CP$ meet $(AIB)$ again at $Q$, so $PAQB$ is harmonic. Now $AC$ is tangent to $(AIB)$, and $AP$ bisects $FD$ (because $AFPD$ is a paralellogram). So $A(P,Q;A,B)$ harmonic, which yields $AQ\parallel DF$. Thus if $X$ is the midpoint of $PQ$, then by homothety $X$ is the intersection of $DF$ and $EG$. But $CX\perp XO$, where $O$ is the center of $(AIB)$ and the midpoint of arc $AB$ on $(CAB)$. Thus $X$ lies on $(CAB)$.

A bary solution: Let $A=(1,0,0),B=(0,1,0),C=(0,0,1),P=(p_1,p_2,p_3)$. We have $a=b$ and it's easy to verify that $P\in (AIB)\iff c^2p_1p_2=a^2p_3^2$. Now $D=(1-p_2,p_2,0),E=(p_1,1-p_1,0),F=(1-p_3, 0, p_3), G=(0,1-p_3,p_3)$. It's easy to verify that the equation of $DF$ is $p_3(1-p_2)y+p_2(1-p_3)z=p_2p_3x$, and that the equation of $EG$ is $p_3(1-p_1)x+p_1(1-p_3)z=p_1p_3y$. Now with Cramer's (the most efficient way is to just ignore the denominators) we can find $X=DF\cap EG=(p_1(p_1+p_2):p_2(p_1+p_2):-p_3^2)$. It's easy to check that $X$ lies on the circumcircle now, given our equation for $P$.

Have you ever tried complex bash??

Let $\{X\}\in FD\cap GE,\{P,Y\}\in FG\cap (AIB)$. Clearly $(AIB)$ is tangent to $AC, BC$ so, by p.o.p. $AF^2=FY\cdot FP$, but $FY=GP$ and the two parallelograms $ADPF, BGPE$ are similar, $\angle AFD=\angle BEG=\angle AEX$, hence $FAXE$ is cyclic; since $FAEP$ is an isosceles trapezoid, $FAXEP$ is cyclic as well and $\angle AXP=\angle AFP\ (\ 1\ )$. In a similar way $BGPDX$ is cyclic and $\angle BXP=\angle AGF\ (\ 2\ )$. Adding $(1)$ and $(2)$ side by side we get $\angle AXB=180^\circ-\angle ACB$, done. Remark: The relations $(1)$ and $(2)$ show also that $C-P-X$ are collinear. Best regards, sunken rock

Remark by symmetry that $AFPE$ and $DPGB$ are isosceles trapezoids, hence they are cyclic. Moreover, since $CF\cdot CA=CG\cdot CB$, $C$ lies on the radical axis of the two circles. Let $H\equiv (AFP) \cap (PGB)$, then $$\angle AHC=\angle AHP=\angle AEP=\angle ABC$$hence $H$ lies on $(ABC)$. Finally, \begin{align*} \angle DHE&=180^{\circ}-\angle AEH-\angle BDH=180^{\circ}-\angle APH-\angle BPH\\ &=180^{\circ}-\angle APB=90^{\circ}-\frac{\angle ACB}{2}=\angle FAE=\angle FHE \end{align*}so that $F, D, H$ are collinear. By symmetry, $G, E, H$ are collinear, so the result is obtained$.\:\blacksquare\:$

Here is my bary solution. Let $P=(x,y,z)$ where $x+y+z=1$. The equation of circumcircle of $AIB$ is :$-a^2yz-b^2zx-c^2xy+(x+y+z)abz=0$. Since $P$ lies on $(AIB) $ we have $a^2z^2=c^2xy$.(here $a=b$.) I will just show how to find $D$ others will be find similarly. Since $D$ lies on $AB$.$D=(t,1-t,0)$.Let the intersection of $PD$ and $CA$ be $Q$.Since $Q$ lies on $AC$ then its coordinates are $(p,0,-p)$ then $u=w$ in $PD$.Then $u(x+z)+vy=0$ $ut+v(1-t)=0$ So $t=\frac{x+z}{x+y+z}$. So $D=(1-y,y,0)$. Similarly we find $F=(1-z,0,z)$ $E=(x,1-x,0)$ $G=(0,1-z,z)$ Intersecting lines $DF$ and $EG$ we get $DF\cap EG=(-x(x+y),y(x+y),z^2)$.We need to show that $a^2z^2(y(x+y)+x(x+y))=c^2xy(x+y)^2$ which is true since $a^2z^2=c^2xy$. $.\:\blacksquare\:$

I am sorry

Ooops...

My solution: We know from Desargue's Theorem on two perspective triangle $PDE,CFG,$ we find $FD,PC,EG$ are concurrent. Let $R=CP\cap (ABC).$ Then we must to prove that $F,D,R$ and $G,E,R$ are collinear. We can find easily $\angle PED=\angle PGC=\angle CFP=\angle PDE\to PD=PE $ where we use parallel lines. We have $\angle CFG=\angle CBA=\angle CRA\to AFPR$ is cyclic.$1.$ Also $\angle CGF=\angle CAB=\angle CRB\to PGBR$ is cyclic.$2.$ Also we know $AFPD$ is parallelogram $\to AF=PD=PE,$ and $FP\parallel AE\to AFPE$ is isosceles cyclic trapezoid.$3.$ Also $EPGB$ is parallelogram $\to GB=PE=PD,$ and $PG\parallel DB\to PGBD$ is isosceles cyclic triangle.$4.$ From $1,3$ we find $AFPER$ is cyclic and from $2,4$ we have $PGBRD$ is cyclic. Also we can find easily $CA$ tangent to $(AIB).$ Then $\angle PRF=\angle FAP=\angle PBA=\angle PED.$ Then $F,D,R$ are collinear, Similarly way we find $R,E,G$ is cyclic.As desired.

Morally correct problem statement wrote: Let $ABC$ be an isosceles triangle with $AB = AC$, whose incentre is $I$. Let $P$ be a point on the circumcircle of the triangle $BIC$ lying inside the triangle $ABC$. The lines through $P$ parallel to $AC$ and $AB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $BC$ meets $AC$ and $AB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC$. [asy][asy] unitsize(100); pair A, B, C, M, P, D, E, F, G, K; A = dir(90); M = dir(270); B = dir(230); C = reflect(A, M) * B; P = M + abs(M - B) * dir(100); K = 2 * foot(origin, A, P) - A; D = extension(P, P + C - A, B, C); E = extension(P, P + B - A, B, C); F = extension(P, P + C - B, A, C); G = extension(P, P + B - C, A, B); draw(F--G, gray(0.5)); draw(P--D^^P--E, gray(0.5)); draw(A--K, gray(0.5)); draw(F--K^^G--K, gray(0.5) + dashed); draw(B--P^^C--P, gray(0.5)); draw(A--B--C--cycle); draw(arc(M, C, B)); draw(circumcircle(B, P, K)^^circumcircle(C, P, K)); draw(circumcircle(B, E, K)^^circumcircle(C, D, K), dotted); draw(unitcircle); dot(A^^B^^C^^P^^D^^E^^F^^G^^K); label("$A$", A, dir(A)); label("$B$", B, dir(200)); label("$C$", C, dir(340)); label("$P$", P, dir(90)); label("$D$", D, dir(290)); label("$E$", E, dir(230)); label("$F$", F, dir(40)); label("$G$", G, dir(140)); label("$K$", K, dir(K)); [/asy][/asy] Let line $AP$ intersect $\odot(ABC)$ again at $K$. We claim that lines $DF$ and $EG$ intersect at $K$. Note that $\overline{AB}$ and $\overline{AC}$ are tangent to $\odot(BPC)$. Since $\angle BKP = \angle BKA = \angle BCA = \angle BDP$, points $B$, $K$, $D$, $P$ are concyclic. Similarly $C$, $K$, $E$, $P$ are concyclic. Since $\angle DPC = \angle PCF = \angle PBC$, $\overline{CP}$ is tangent to $\odot(BDP)$. Similarly, $\overline{BP}$ is tangent to $\odot(CEP)$. Since $\angle KDC = 180^{\circ} - \angle KDB = 180^{\circ} - \angle KPB = 180^{\circ} - \angle KCP$, $\overline{CP}$ is also tangent to $\odot(KDC)$. Thus line $KD$ bisects $\overline{CP}$; since $CDPF$ is a parallelogram, $\overline{DF}$ also bisects $\overline{CP}$, so $K$, $D$, $F$ are collinear. Similarly $K$, $E$, $G$ are collinear as desired.

Here's another approach: Invert about $P$ with radius $\sqrt{PB \cdot PC}$ followed by reflection in the angle bisector of $\angle BPC$. Then we get the following problem (as the above user says, we correct the restatement morally ):- Inverted problem wrote: Let $H_A$ be the $A$-Humpty point of $\triangle ABC$, and let $\omega$ be its circumcircle. Suppose the tangent to $\omega$ at $A$ meets $\odot (AH_AB) \equiv \gamma_B$ and $\odot (AH_AC) \equiv \gamma_C$ at $F$ and $G$ respectively. Also let us assume that the tangents to $\gamma_B$ and $\gamma_C$ at $A$ meet $\omega$ at $D$ and $E$ respectively. Show that $\odot (ADF),\odot (AEG),\odot (BH_AC)$ meet at a point. Let $BF \cap CG=K$. We show that $K$ is the desired point. Now, $$\angle BFG=\angle CBA=\angle GAC \Rightarrow AC \parallel BF$$Also $AE \cap BC$ lies on the perpendicular bisector of $AC$ (As this is the point where the tangents to $\gamma_C$ at $A$ and $C$ meet). This gives that $BE \parallel AC$, i.e. $F,B,E$ are collinear. And, $\angle GDA=\angle CBA=\angle GFK$, which means that $K \in \odot (ADF)$. Similarly, $K \in \odot (AEG)$. Also, as $BK \parallel AC$ and $CK \parallel AB$, so $ABKC$ is a parallelogram. Thus, $$\angle BKC=\angle BAC=180^{\circ}-\angle BH_AC \Rightarrow K \in \odot (BH_AC) \quad \blacksquare$$

Since, there are almost six elements passing through a single point, there are many methods to approach this problem...I'll post another one ISL 2003 G5 wrote: Let $ABC$ be an isosceles triangle with $AC=BC$, whose incentre is $I$. Let $P$ be a point on the circumcircle of the triangle $AIB$ lying inside the triangle $ABC$. The lines through $P$ parallel to $CA$ and $CB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $AB$ meets $CA$ and $CB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC$. Solution: $\angle ABC=\angle CAB=\angle PDB \implies PDGB$ is cyclic. Similarly, $AFPE$ is cyclic and obviously $AFGB$ is cyclic. Let $\odot (ABC)$ $\cap$ $\odot (AFPE)$ $=$ $M$, Let $MP \cap \odot (ABC) = C'$ $$\angle AFP=180^{\circ}-\angle ABC=180^{\circ}-\angle AMC'=180^{\circ}-\angle ABC' \implies C' \equiv C \implies C - P - M$$$\angle KMB=\angle CAB=\angle CGF \implies PGDMB$ is cyclic. $\angle CAI=\angle IBA$ $\implies$ $CA , CB$ tangent to $\odot (AIPB)$, hence, $\angle CAP$ $=$ $\angle APD$ $=$ $\angle PBA $ $\implies$ $AP$ is tangent to $\odot (PGDMB)$ and similarly, $BP$ is tangent to $\odot (AFPEM)$ $\implies$ $\angle AFM$ $=$ $\angle APM$ $=$ $\angle FGM$ $\implies$ $CF, CG$ are tangents to $\odot (FGM)$, Now $$\angle MGB=\angle MDB=\angle MFG \implies F - D - M \text{ and similarly, } E - G - M$$

It's obvious that lines $FD$,$GE$,$CP$ passes through one point. Now we denote by $K$ the intersection of the circumcircle of $ABC$ and the line $CP$. We will show that $K$ also lies on lines $EG$ and $FD$. First,note that the quadrilaterals $PFAK$ and $BKPG$ are cyclic. So we see that pentagon $BGPDK$ is cyclic too. Thus we have $$\angle FKP=\angle PAC,\angle PKD=\angle PBA$$. Now note that $\angle CAI=\angle IBA$, thus $CA$ is tangent to $(ABI)$. But it means that $\angle PKF=\angle PKD$. So points $F$,$D$ and $Q$ are collinear$.\:\blacksquare\:$

I'm trying to find a different solution. Here is what I currently have and I would appreciate it if someone knows how to finish the solution. By Desargues theorem $DF, EG, CP$ concur at a point, call it $X'$, and let $X=CP \cap (ABC)$. Futhermore, let $CM \cap \odot (ABC)=M_C$, and $M_CX \cap FG=L$, and $MP \cap CL=Y$ where $M$ is the midpoint of $AB$. Then, obviously, $L$ is the orthocenter of $\triangle APM_C$, so $\angle CLM_C=\pi-\angle APM_C=\pi-\angle PMM_C=\angle CML$ since $$M_CM \times M_CC =MI^2=MP^2 \implies \triangle PMM_C ~\triangle CPM_C$$ so $MM_CLY$ is cyclic. Therefore, it suffices to show given $M_CX' \cap FG=L'$, then $MM_CL'Y$ is cyclic. I got stuck here, so if anyone has any ideas, that would be awesome.

Oh what, this is G5? Hmm, lol, I thought it was just a Singapore TST 2004 problem. This is very easy for G5? in my opinion. Let $K=(ABC)\cap CP$, I claim that $K$ lies on $DF$ and $EG$. Let $Q=(BIC)\cap GF$. Claim. $PFAKE$ is cyclic. $$\measuredangle AKP=\measuredangle AKC=\measuredangle ABC=\measuredangle CAB=\measuredangle FAB=\measuredangle AFP$$$$\measuredangle AEP=\measuredangle ABC=\measuredangle CAB=\measuredangle FAB=\measuredangle AFP$$ Claim. $BGPDK$ is cyclic. $$\measuredangle PKB=\measuredangle CKB=\measuredangle CAB=\measuredangle ABC=\measuredangle ABG=\measuredangle PGB$$$$\measuredangle PDB=\measuredangle CAB=\measuredangle ABC=\measuredangle ABG=\measuredangle PGB$$ Claim. $\triangle PEG\sim \triangle PFD$. Since $Q,P$ lie on $(BIC)$ and it is well-known that centre of $(BIC)$ is the midpoint of arc $AB$, we have that $BC$ is tangent to $(BIC)$ and therefore by PoP, $$GB^{2}=GQ\cdot GP\implies \triangle BGQ\sim \triangle PGB.$$Hence, we have $$\frac{PG}{PE}=\frac{PG}{GB}=\frac{GB}{GQ}=\frac{PD}{PF},$$since $BG=PE=PD$ and $GQ=PF$. Also, easy to see that $\angle GPE=\angle DPF$. These can be obtained easily by some parallelogram and trapezoid properties. Now claim follows. By the last claim, we have a spiral similarity with centre $P$ taking $DF$ to $EG$, thus there is also a spiral similarity with centre $P$ taking $FE$ to $DG$. Therefore $DF,EG,(PEF)$ and $(PGD)$ all concur and since $K$ lies on $(PEF)$ and $(GPD)$, we conclude that indeed $DF,EG$ intersect on the $(ABC)$.