Circle $ \mathcal C$ with center $ C$ and radius $ CG$ cuts the rays $ AB, AD, AF$ at $ B', D', F'.$ $ B'D'GF'$ is cyclic isosceles trapezoid. Since $ CB \perp B'G$ and $ CD \perp F'D',$ $ B, D$ are midpoints of its diagonals $ B'G, F'D',$ $ BD$ is its midline. It cuts the segment $ D'G$ at its midpoint $ M,$ so that $ DM$ is the D-median of $ \triangle DD'G.$ $ CM$ is perpendicular bisector of the chord $ D'G$ $ \Longrightarrow$ $ M$ is on a circle $ \mathcal K$ with diameter $ CG.$ The angles $ \angle CBG, \angle CHG$ are right $ \Longrightarrow$ $ B, H \in \mathcal K.$ Parallel to $ DA$ through $ B$ cuts $ CG$ at $ P.$ $ BPDF$ is a parallelogram, its diagonals $ BD, PF$ cutting each other in half at $ E,$ so that $ BE$ is the B-median of $ \triangle BPF.$ Triangles $ \triangle BPF \sim \triangle DD'G$ are centrally similar, having parallel 2 pairs of corresponding sides $ BF \parallel DG, BP \parallel DD'$ and corresponding medians $ BE \parallel DM$ $ \Longrightarrow$ $ EF \parallel MG.$ Since quadrilateral $ BMGH$ is cyclic with circumcircle $ \mathcal K,$ quadrilateral $ BEFH$ with equal internal angles is also cyclic.

$ BF||DG\rightarrow \frac{AF}{AD}=\frac{BA}{AG}\leftrightarrow AF.AG=AB^2=AE.AC$ $ \rightarrow \frac{AF}{AE}=\frac{AC}{AG}\rightarrow \triangle AFE \sim \triangle ACG$ $ \rightarrow \angle{AFE}=\angle{ACG}$ $ \angle{CHF}=\angle{FDC}=\angle{CBG}=90$ $ CHFD,CBHG$ are cyclic $ \rightarrow \angle{GFE}=\angle{GFD}+\angle{AFE}=\angle{HCD}+\angle{ACG}$ $ =\angle{HCG}+\angle{ACD}=\angle{HBG}+\angle{ABD}=\angle{ABE}$ =>$ HFEB$ is cyclic

Let $CH$ meet $BD$ at $I$, and let $CE$ meet $FG$ at $J$. $FHEB$ cyclic is equivalent to $JI || FB$, since $\angle HJI = \angle HEI$, and $\angle BFH = 180^{\circ} - \angle HEI$ if and only if $BFHE$ is cyclic and if and only if $JI || BF$. Since $BF || GD$, to show $IJ || BF$ it suffices to show $\frac{FJ}{JG} = \frac{BI}{ID}$. Since $\angle FHC = \angle FDC = 90^{\circ}$, $FHDC$ is cyclic, and since $\angle GHC = \angle GBC = 90^{\circ}$, $GHBC$ is cyclic. Therefore, $\angle AFG = \angle ICD$ and $\angle AGF = \angle BCI$. In addition, since $CA$ bisects $\angle FAG$, we have $\frac{FA}{AG} = \frac{FJ}{JG}$. Thus, \[ \frac{FJ}{JG} = \frac{FA}{AG} = \frac{\sin \angle AGF}{\sin \angle AFG} = \frac{\sin \angle BCI}{\sin \angle ICD} = \frac{CB \cdot CI \sin \angle BCI}{CD \cdot CI \sin \angle ICD} = \frac{2[BCI]}{2[ICD]} = \frac{BI}{ID}, \] as desired.

Good problem , i have same solution as mitdac123

$\frac{AF}{AD}=\frac{AB}{AG} \Rightarrow AB^2=AF \cdot AG=AE \cdot AC \Rightarrow \frac{AE}{AF}=\frac{AG}{AC} \Rightarrow \triangle AFE \sim \triangle ACG$, and quad $BHGC$ is cyclic then problem solved, I don't know why in the given figure $BHGC$ is not cyclic...

WRONG SOLUTION: We know $BHGC$$(1)$ is cyclic,then $$\angle BHG=90+\angle BHC=^{(1)} 90+\angle BGC=90+\angle AGD=90+(90-\angle DAG)=180-\angle DAG=180-\angle BAF\to \angle BHG+\angle BAF=180,$$then $BHFA$ is cyclic.Also we know $\angle BFA=\angle BEA=90\to BFAE$ is cyclic.Then $BHFAE$ is cyclic.As desired.

Ferid.---. wrote: My easy solution: We know $BHGC$$(1)$ is cyclic,then $$\angle BHE=90+\angle BHC=^{(1)} 90+\angle BGC=90+\angle AGD=90+(90-\angle DAG)=180-\angle DAG=180-\angle BAE\to \angle BHE+\angle BAE=180,$$then $BHEA$ is cyclic. How can you tell that $BHEA$ is cyclic? $B , E , F$ and $H$ are to be proved cyclic not $BHEA$.

RC. wrote: Ferid.---. wrote: My easy solution: We know $BHGC$$(1)$ is cyclic,then $$\angle BHE=90+\angle BHC=^{(1)} 90+\angle BGC=90+\angle AGD=90+(90-\angle DAG)=180-\angle DAG=180-\angle BAE\to \angle BHE+\angle BAE=180,$$then $BHEA$ is cyclic. How can you tell that $BHEA$ is cyclic? $B , E , F$ and $H$ are to be proved cyclic not $BHEA$. I am sorry,it is my mistake,and I write full and true solution in the last post.

Ferid.---. wrote: RC. wrote: Ferid.---. wrote: My easy solution: We know $BHGC$$(1)$ is cyclic,then $$\angle BHE=90+\angle BHC=^{(1)} 90+\angle BGC=90+\angle AGD=90+(90-\angle DAG)=180-\angle DAG=180-\angle BAE\to \angle BHE+\angle BAE=180,$$then $BHEA$ is cyclic. How can you tell that $BHEA$ is cyclic? $B , E , F$ and $H$ are to be proved cyclic not $BHEA$. I am sorry,it is my mistake,and I write full and true solution in the last post. $\angle BGC \not =\angle AGD$ I'm afraid

CeuAzul wrote: Ferid.---. wrote: Ferid.---. wrote: My easy solution: We know $BHGC$$(1)$ is cyclic,then $$\angle BHE=90+\angle BHC=^{(1)} 90+\angle BGC=90+\angle AGD=90+(90-\angle DAG)=180-\angle DAG=180-\angle BAE\to \angle BHE+\angle BAE=180,$$then $BHEA$ is cyclic. How can you tell that $BHEA$ is cyclic? $B , E , F$ and $H$ are to be proved cyclic not $BHEA$. I am sorry,it is my mistake,and I write full and true solution in the last post. $\angle BGC \not =\angle AGD$ I'm afraid That's exactly what I was trying to say. I have a simple proof using similarity which I will add tomorrow as it is IMO today.

RC. wrote: CeuAzul wrote: Ferid.---. wrote: My easy solution: We know $BHGC$$(1)$ is cyclic,then $$\angle BHE=90+\angle BHC=^{(1)} 90+\angle BGC=90+\angle AGD=90+(90-\angle DAG)=180-\angle DAG=180-\angle BAE\to \angle BHE+\angle BAE=180,$$then $BHEA$ is cyclic. How can you tell that $BHEA$ is cyclic? $B , E , F$ and $H$ are to be proved cyclic not $BHEA$. I am sorry,it is my mistake,and I write full and true solution in the last post. $\angle BGC \not =\angle AGD$ I'm afraid That's exactly what I was trying to say. I have a simple proof using similarity which I will add tomorrow as it is IMO today. I am sorry ,I don't understand the condition of question.I think $G$ lies on $CD.$ Then my solution is wrong.I am sorry.

Since, $AC$ is a diameter of \(\odot ABCD\) and \(AC\perp BD\) we know obviously that \(ABCD\) is a kite with \(AB = AD\) and also \(\angle BAC = \angle CAD => \angle EAF = \angle GAC - (1)\) Since, \(DG || BF => \dfrac{AF}{AD}=\dfrac{AB}{AG} => AB\times AD = AF\times AG.\) In right \(\Delta ABC\) \(AB^{2} = AE * AC => AB \times AD = AE \times AC => AF \times AG = AE\times AC =>\dfrac{AF}{AE}=\dfrac{AG}{AD}\) and from [1] \(\angle EAF = \angle GAC => \Delta EAF \sim \Delta GAC => \angle FEA = \angle AGC = \angle BHC.\) \(\angle GHB + \angle FEB = \angle GHC +\angle BEA = 180^{\circ}\)

mitdac123 wrote: $ BF||DG\rightarrow \frac{AF}{AD}=\frac{BA}{AG}\leftrightarrow AF.AG=AB^2=AE.AC$ $ \rightarrow \frac{AF}{AE}=\frac{AC}{AG}\rightarrow \triangle AFE \sim \triangle ACG$ $ \rightarrow \angle{AFE}=\angle{ACG}$ $ \angle{CHF}=\angle{FDC}=\angle{CBG}=90$ $ CHFD,CBHG$ are cyclic $ \rightarrow \angle{GFE}=\angle{GFD}+\angle{AFE}=\angle{HCD}+\angle{ACG}$ $ =\angle{HCG}+\angle{ACD}=\angle{HBG}+\angle{ABD}=\angle{ABE}$ =>$ HFEB$ is cyclic How did you get the first line? Where did you get that $AF\cdot AG=AB^2=AE\cdot AC$

Since $\angle{FHD}+\angle{FCD}=180$ So $GF.GH=GC.GD=GA.GB$ and $\angle{AFB}+\angle{AEB}=180$ then HFAEB is cyclic