Easy problem notice that $BF$ is tangent to the $O$ and $\angle OGF=OBF=90^{\circ}$ . let $BD\cap AE=X$ , $CF$ is polar of $X$ and we know that $OX$ is perpendicular to polar of $X$.hence $O,G,X$ are collinear.hence $XB.XD=XG.XO=XE.XA$

Why is $ CF$ a polar of $ X$? I don't see how we can deduce that from what we know.

hollandman wrote: Why is $ CF$ a polar of $ X$? I don't see how we can deduce that from what we know. Cause X lies on polar of F wrt (O) .

Amir.S wrote: hollandman wrote: Why is $ CF$ a polar of $ X$? I don't see how we can deduce that from what we know. Cause X lies on polar of F wrt (O) . Sorry if I'm missing something obvious, but how do we know $ X$ lies on the polar of $ F$?

Because FB and FD are tangent to (O)

I have a new way :notice that:∠BEC=∠BAD;∠BGF=∠BOF; Obviously,OF ∥AD; ∠BOF=∠BAD; So we get:∠BEC=∠FGB; SO :G,B,E,C are concyclic.Hence:∠GEB=∠GCO; Notice that :∠AEG+∠GEB=90;∠GCO+∠GOC=90; SO:∠AEG=∠GOC..The answer easily follows .

Amir.S wrote: hollandman wrote: Why is $ CF$ a polar of $ X$? I don't see how we can deduce that from what we know. Cause X lies on polar of F wrt (O) . Every point is determined by 2 lines, and every line is determined by 2 points. You only have 1 line (p(F)) that goes through X, which only implies that F lies on the polar of X, but there are infinitely many such lines.

We will prove that if $G$ is the intersecion of the circumcircles of $AEO$ and $BOD$ then we have $G-F-C$ I have a proof by inversion. Invert WRT circle $O$. $A,E,D,B$ stay in the same place. We have to prove that $G'OC'F'$ is cyclic. We see that $F'$ lies on $BD$ so $F'$ is midpoint of $BD$. Clearly $G' \equiv AE \cap BD$ and $OC'DE$ is cyclic. Clearly, since $AD$ and $BE$ are altitudes of triangle $ABG'$ ,the circumcircle of triangle $ODE$ is the nine point circle of triangle $AG'B$ so $C'$ is the feet of the altitude from $G'$ to $AB$. Hence $\angle OC'G' = 90 = \angle OF'D$ and therefore $G'OC'F'$ is cyclic so we are done.

Let $AE\cap BD = M$ and $AD \cap BE = N$, we have $AB \cap ED = C$, so by Brocard's theorem $CN$ is polar of $M$ with respect to the circle centered at $O$. Let $MO \cap QC = T$, so $\angle MTF = \angle OTF= \frac{\pi}{2}$, so $T$ lies in the circle with diameter $OF$. $\angle MEN = \angle MDN= \frac{\pi}{2}$. Now $\angle MED= \angle DBO = \angle MTD$.so $M,D,N,T,E$ are cyclic with $MN$ as diameter. so $\angle MQN= \angle MQF = \angle MQC = \frac{\pi}{2}$. So $C,F,N$ are collinear and hence $T= G$. So $M,G,O$ are collinear. By intersecting chord theorem $ME \cdot MA = MD\cdot MB = MG \cdot MO$, so $A,E,G,O$ are concyclic.

Dear Mathlinkers, another way uses the pivot theorem three times Sincerely Jean-Louis

Use complex numbers; let all lowercase letters denote the complex number of that point in the problem. Set $a=-1, b=1$. We can compute using the circumcenter formula that the circumcenter of $\triangle BOD=\frac{d}{1+d}$, whence $F=\frac{2d}{1+d}$, and the circumcenter of $\triangle AOE= \frac{e}{1-e}$. Note that $G$ is the reflection of $O$ across the line formed by the circumcenters of $\triangle AOE$ and $\triangle BOD$, so reflecting $O$ over the line formed by the circumcenters gives $G=\frac{d+e}{d-e+2}$. Thus, we check that $$\frac{\frac{2d}{1+d}-\frac{d+e}{de+1}}{\frac{d+e}{de+1}-\frac{d+e}{d-e+2}} \in \mathbb{R}$$$$\frac{(d-1)(d-e+2)(d-e-2ed)}{(e-1)(d+1)^2(d+e)}\in \mathbb{R}$$which is a straightforward, albeit tedious computation.

Extension: Prove, that, Points $B,E,G,C$ are also concyclic

AlastorMoody wrote: Extension: Prove, that, Points $B,E,G,C$ are also concyclic Note that $\angle{BEC}=\angle{BAD};\angle{BGF}=\angle{BOF}.$ Also, $OF \parallel AD$ and so $\angle{BOF}=\angle{BAD}$. So we get $\angle{BEC}=\angle{FGB}$. Thus $B,E,G,C$ are concyclic. REMARK: Proceeding in this manner also gives us a solution. By our observation, $\angle{GEB}=\angle{GCO}$ Note that $\angle{AEG} + \angle{GEB}=90$ and $\angle{GCO}+\angle{GOC}=90$ Thus the desired conclusion follows.

Infact using the extension, the original problem becomes 2 liner Solution:

, we know that $GCBE$ is cyclic, hence, Let $\angle EAB=x \implies \angle EBA=90^{\circ}-x \text{ and } \angle EGC =90^{\circ}+x $, obviously, $$\angle OGF=90^{\circ} \implies \angle OGE=360^{\circ}-\left( 180^{\circ}+x \right) =180^{\circ}-x=180^{\circ}-\angle EAB$$

So we shall do an inversion $\psi(\Gamma_{ADBE})$ which uses the circle around $ADBE$, so by this inversion we have $\gamma \xrightarrow{\psi} BD$,here we are talking about line $BD$,not the segment $A \xrightarrow{\psi} A$,the same is for points $E,B,D$,and we have that $G \xrightarrow{\psi} G'$ So if the quad $AGOE$ is cyclic then, we have that the points $G',A$ and $E$,that is because of the following: $$180 = \angle OGA + \angle OEA = \angle OA'G' + \angle OA'E' = \angle OAG' + \angle OAE$$Here we have that $A \equiv A'$,and $E \equiv E'$ Thus we need to show that the points $G',A$ and $E$ are concylic. Now if that condition is true then we must have: $$G'D.G'B=G'A.G'E$$$$\frac{GD}{OG.OD}r^2.\frac{GB}{OG.OB}r^2 = \frac{GA}{OG.OA}r^2.\frac{GE}{OG.OE}r^2$$$$GD.GB=GA.GE$$$$\frac{GA}{GB} = \frac{GD}{GE}$$Thus we need to show the following relation $\triangle GAB \sim \triangle GDE$,now what that implies is that the quads $ACGD$ and $ECGB$ are concyclic. This is easily seen by a quick angle-chase: $$\angle CEB = \angle DEB = \angle BDF = \angle BGF $$$$ \angle CAD = \angle BAD = \angle DBF = \angle DGF $$This is due to the fact that the lines $FD,FB$ are tangents to $\Gamma$,because of the right angles $\angle ODF = \angle OBF = 90$ Thus we have proven that these two are concylic. Thus the ratio holds. Thus the points $G',A,E$ are all collinear. Thus the quad $AGOE$ is cyclic...

Let $AD\cap BE=H$ and $AE\cap BD=J$. By brocard's theorem $CH$ is polar of $J$. Let $J^*$ be the inverse image of $J$ under the inversion about $\Gamma$. So, $J^*$ is miquel point of $ABED$ and $\angle OJ^*H=90^\circ$ and $J^*\in CH$. Now, $\angle FDO=180^\circ-\angle FBD=90^\circ$. So, $BD$ is polar of $F$ and $J\in BD$ and thus, by "La'Hire" theorem $F\in\text{polar of } J\equiv CH$. So, $$\angle OJ^*F=\angle OJ^*H=90^\circ\implies J^*\odot(BOD)\implies J^*\equiv G.$$So, $G$ is the miquel point of $ABED$. Since, $ABED$ is cyclic we have $G\in\odot(AOE)$.

We invert wrt to the circle with diameter $\overline{AB}$. Notice that since $(DBOG)$, we have that the inverse of $G$, $G'$ say, is $OG\cap DB$. With this inversion, the four points $O,E,A,G$ are respectively, $P_{\infty},E,A,G'$, thus we proceed to show that $G'-E-A$ collinear, which would imply our result. By angle chasing, we have $$\angle DGF = \angle DBF = \frac{180^\circ-\angle DFB}{2} = \frac{\angle DOB}{2} = \angle DAB \Rightarrow (DACG)$$similarly, we have $(BECG)$, which together gives us $$\angle AGC = \angle ADC = \angle ABE = \angle CGE$$Furthermore, since $FC$ is the diameter of circle $(DFBO)$, we have $FG\perp OG$, which implies that $\angle G'GA = \angle OGE = 90^\circ- \angle ADE$. By noticing that $OG'\cdot OG = OA^2 = OB^2$, we have $\Delta OGA\sim \Delta OAG'$, and $\Delta OGE\sim\Delta OEG'$. Thus $$\angle OAG'= 90^\circ + \angle ADE$$However, in the triangle $\Delta OAB$, notice that $\angle OAB = \frac{180^\circ - \angle AOB}{2} = 90^\circ-\angle ADE$. These two arguments imply that $G'-A-E$ and we are done.

Let $AE \cap BD=P , AD \cap BE=R$ By Brocard's Theorem: $\triangle PRC$ is autopolar $ \implies p \equiv PC$ Then $OP \perp RC$ Let $G^*=OP \cap RC$ but $\angle PG^*R = \angle PER = \angle RDP$ $\implies P,E,G^*,R,D$ are concyclic. Then: $\angle PG^*D= \angle PED=\angle DBA$ $\implies O,B,D,G^*$ are concyclis $\implies G^*=G$ By power of a point in P: $PG.PO=PD.PB$ and $PE.PA=PD.PB$ $\implies PE.PA=PG.PO$ $\implies O,A,E,G$ are concyclics

Claim: $BCGE$ is cyclic. Proof : First Note that $\angle ODF = \angle 90$ so $FD$ is tangent to $O$. $\angle BEC = \angle BAD = \angle BDF = \angle BGF = \angle BGC$. we want to prove $\angle OGF + \angle FGB + \angle BGE - \angle OAE = \angle 180$. $\angle OGF = \angle 90$, $\angle FGB = \angle DAB$, $\angle BGE = \angle BCE$ and $\angle OAE = \angle BDC$ so $\angle OGF + \angle FGB + \angle BGE - \angle OAE = \angle 90 + \angle DAB + \angle DBC = \angle 180$. we're Done.

By symmetry, $F$ is the arc midpoint of $\widehat{BD}$, so $\overline{OF} \perp \overline{BD}$ and thus $\overline{OF} \parallel \overline{AD}$. Therefore, $$\measuredangle CEB=\measuredangle DEB=\measuredangle DAB=\measuredangle FOB=\measuredangle FGB=\measuredangle CGB,$$hence $CBEG$ is cyclic. We now present two approaches. Approach 1 (angle chase): Observe that $$\measuredangle AEG=90^\circ-\measuredangle GEB=90^\circ\measuredangle GCB=90^\circ=\measuredangle GCO=\measuredangle COG=\measuredangle AOG,$$since $\angle AEB=\angle OGF=90^\circ$, so we are done. $\blacksquare$ Approach 2 (inversion): Invert about $(ADBE)$, sending every point $P$ to a point $P'$, so $F'$ is the foot of the perpendicular from $O$ to $\overline{BD}$. Since $F,C,G$ are collinear, $OF'C'G'$ is cyclic, so $\angle OC'G'=\angle BC'G'=90^\circ$. Since $C'G'BE$ is still cyclic, this means $\angle BEG'=90^\circ$, but we also have $\angle BEA=90^\circ$, hence $A,E,G'$ are collinear, so by inverting back (which fixes $A,E$) we find that $AOGE$ is cyclic, as desired. $\blacksquare$

we can use inversion