My solution: Since $\angle ACB=90^\circ,$ so $\overline{AB}$ is diameter of circle $O_2.$ Since $\triangle AEF$ is similar to $\triangle CEB,$ so \[\frac{AH}{HF}=\frac{AE\cdot\sin{\angle AEH}}{FE\cdot\sin{\angle FEH}}=\frac{EC}{EB}\cdot\frac{BG}{GC}.\qquad (1)\] On the other hand, \[\frac{AC}{CD}=\frac{AC}{BC}\cdot \frac{BC}{CD}=\frac{EC}{EB}\cdot \frac{BG}{GC}\qquad (2)\]So done. $\blacksquare$ Remark: You may find $F,H,C,G$ are concyclic or something else, but they're inessential. [asy][asy] size(8cm); pathpen=black; pointpen=black; pointfontpen=fontsize(9pt); void b(){ pair O1=D("O_1",origin,dir(70)); pair B=D("B",dir(180),dir(-135)); pair C=D("C",dir(0),dir(-45)); pair O2=D("O_2",(0,1.5),dir(-10)); pair A=D("A",IP(CP(O2,C),L(B,O2,-0.01,5)),dir(90)); pair E=D("E",IP(D(A--B),unitcircle),dir(170)*3); pair F=D("F",IP(L(C,E,-0.01,5),CP(O2,B)),dir(180)); pair H=D("H",2*F/3+A/3,dir(100)); pair G=D("G",OP(L(H,E,-0.01,5),unitcircle),dir(-100)); pair D=D("D",extension(B,G,A,C),dir(-90)); D(unitcircle,magenta+dashed); D(CP(O2,B),magenta+dashed); D(C--F); D(A--F); D(B--D); D(A--D); D(H--G); D(B--C);D(C--G); D(anglemark(A,E,H,7),red); D(anglemark(B,E,G,7),red); D(anglemark(H,E,F,7),blue); D(anglemark(G,E,C,7),blue); } b(); pathflag=false; b(); [/asy][/asy]

@Complex2Liu Are you assuming that $\angle{FHE} = 90^{\circ}$ because how did you get that $\frac{AH}{HF}=\frac{AE\cdot\sin{\angle AEH}}{FE\cdot\sin{\angle FEH}}$?

My solution: Completing angles we get $\angle CBA=\angle CFA=\angle FCA=\theta$ and $\angle GEC=\angle GBC=\angle FEH=\beta$. Let $X$ be a point in $CF$ such that $XA\parallel EH$ and let $Y$ in $ AE$ such that $XY\parallel FA$ $\Longrightarrow$ $\angle XAY=\angle BDA=90^{\circ}-\beta$ and $\angle XYA=\angle BAD=90^{\circ}-\theta$ $\Longrightarrow$ $\triangle BDA\cup (C)\sim \triangle XAY\cup (E)$ $\Longrightarrow$ $\tfrac{YE}{EA}=\tfrac{AC}{CD}...(\star)$, but since $XA\parallel EH$ and $XY\parallel FA$ we get $\tfrac{YE}{EA}=\tfrac{XE}{EF}=\tfrac{AH}{HF}...(\star \star)$. Combining $(\star)$ and $(\star \star)$ we get $\tfrac{AH}{HF}=\tfrac{AC}{CD}$.

CGMO 2002, Problem 4

$\angle BGE = \angle BCE = \angle BAF \Rightarrow$ $A, D, G, E$ concyclic. Then \[\frac{AH}{HE} = \frac{AE}{EF}\frac{\sin \angle BEG}{\sin \angle GEC} = \frac{AE}{EF}\frac{BG}{BC}.\]and \[\frac{AC}{CD} = \frac{AB}{BD} \frac{\sin \angle CBE}{\sin \angle DBC} = \frac{AB}{BC}\frac{CE}{CG}.\]These two are equal iff \[\frac{AE}{EF} BC = \frac{AB}{BD} CE \Leftrightarrow \frac{CE}{BE} BG = \frac{AB}{BD} CE \Leftrightarrow \frac{BF}{BE} = \frac{AB}{BD},\]which is true because $A, D, G, E$ are concyclic.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Highway%20to%20Geometry%204.pdf p. 20... Sincerely Jean-Louis

Let $r_1,r_2$ be the radii of $O_1,O_2$ respectively. Note that $\overline{BC} \perp \overline{ACD}$ and $\overline{CE}\perp \overline{AB}$. We have \[ \tan \angle FEH = \tan \angle GEC = \tan \angle DBC = \frac{CD}{2r_1}\]so $CD=2r_1\tan \angle FEH$. Let $\theta=\angle CAB = \angle BCE$. Then $AC=\tfrac{2r_1}{\tan \theta}$, so $\tfrac{CD}{AC}=\tan \theta \tan \angle FEH$. By the Ratio Lemma in $\triangle AEF$, we have \[ \frac{HF}{HA} = \frac{EF\sin \angle FEH}{EA\sin \angle AEH} = \frac{EF}{EA}\cdot \tan \angle FEH.\]since $\angle AEF=90$. But $\angle EAF=\angle BCE$, so $EF/EA = BE/EC=\tan \theta$. Therefore, $\tfrac{HF}{HA} = \tan \theta \tan \angle FEH$, and we are done.

By Reim theorem, the line tangents to $O_1$ at $E$ is parallel to $AF$ Hence, $$\frac{AC}{CD}=(A,D;C,\infty_{AC})=(E,G;C,B)=(B,C;G,E)=(A,F;H,\infty_{AF})=\frac{AF}{FH}$$

Move $H$ linearly along $AF$. Then $D$ also moves linearly. Checking two cases we see that the desired ratio holds.