We have: $ \frac{AH}{HF}=\frac{S_{AEH}}{S_{FEH}}=\frac{AE.\sin\widehat{{AEH}}}{FE.\sin\widehat{{FEH}}}=\frac{CE.\sin\widehat{{BEG}}}{BE.\sin\widehat{{CEG}}}=\frac{CE.BG}{BE.CG}$ And: $ \frac{AC}{CD}=\frac{S_{ABC}}{S_{DBC}}=\frac{AB.CE}{DB.CG}$ Combine with $ BE.BA=BC^2=BG.BD\Leftrightarrow \frac{AB}{DB}=\frac{BG}{BE}$ we have solution.

This is a nice solution which a friend of mine has come up with today. Since $AC$ is tangent to $\odot O_2$, $\angle BCA=90^\circ$, so as $\angle AFB$. Moreover, we also have $\angle CEB = 90^\circ$ for $BC$ is a diametre of $\odot O_1$. So, $\angle HAE =\angle BAD$. On the other hand, $ \angle AHE$ $= 180^\circ - \angle FAB - \angle HEA$ $= 180^\circ - \angle ECB - \angle BEG$ $= 180^\circ - \angle EGB - \angle BEG$ $= \angle ABD$ which leads to $\triangle AHE \sim \triangle ABD$. Hence $\frac{AH}{AB}=\frac{AE}{AD}$. Then obviously $\triangle AEF \sim \triangle ACB$, giving that $\frac{AB}{AF}=\frac{AC}{AE}$. Combining the two equalities above we get $\frac{AH}{AF}=\frac{AC}{AD}$, and the result follows.

I think this is very similar to Uagu 's solution Observe that 1)$ACBF$ is a kite 2) Hence $\Delta AHE \sim \Delta ABD$ $\Rightarrow \frac{AH}{AE}=\frac{AB}{AD} \Rightarrow AH.AD=AE.AB =AC^2 =AC.AF \Rightarrow \frac{AH}{AF}=\frac{AC}{AD}$ Done!!

My solution: We know $\angle BCA=90,\angle FEA=90,\angle EAF=\angle BAC,$ then $\triangle BCA\sim \triangle FEA\to \frac{EA}{EF}=\frac{CA}{CB}(1).$ Let $\angle EFA=x,\angle FEH=y,$(Also $\angle DBC=\angle GEC=\angle FEH=y.$) then we know $$\frac{AH}{HF}=\frac{tg(x)}{tg(y)}=\frac{EA}{EF}\cdot \frac{BC}{CD}=^{(1)}\frac{AC}{BC}\cdot \frac{BC}{CD}=\frac{AC}{CD}.$$$Q.E.D$

This is easy computation. Let $\angle HAE= \alpha, \angle AEH= \beta$. Let $R$ denote radius of circle $O_2$. Note that $\angle BCA=\angle AEC=\angle BGC=90^{\circ}$ By direct angle chasing, we get $\angle BAC=\alpha$ and $\angle BDC=\beta$, which gives $AC=2R \cos \alpha$ and $CD=2R \sin \alpha \cot \beta$ $$\frac{AC}{CD}=\cot \alpha \tan \beta$$ Now we compute $AH$ and $HF$. By sine rule on $\Delta AHE$, \begin{align*} AH &= \sin \angle HEA \cdot \frac{HE}{\sin \angle HAE}\\ &=\sin \beta \cdot \frac{HE}{\sin \alpha}\\ \end{align*}By sine rule on $\Delta FHE$, \begin{align*} HF&= \sin \angle FEH \cdot \frac{HE}{\sin \angle HFE} \\ &=\cos \beta \cdot \frac{HE}{\cos \alpha} \\ \end{align*} \begin{align*} \implies \frac{AH}{HF} &=\frac{\sin \beta \cdot \frac{HE}{\sin \alpha}}{\cos \beta \cdot \frac{HE}{\cos \alpha}}\\ &=\tan \beta \cot \alpha\\ \end{align*}As desired.

I want to attach a diagram but don' know how to do it. anybody help?

tdl wrote: We have: $ \frac{CE.\sin\widehat{{BEG}}}{BE.\sin\widehat{{CEG}}}=\frac{CE.BG}{BE.CG}$ how we get it?

Hmm don't these solutions assume BECG is cyclic, which clearly isn't true because BEC are on O1 and G isn't?

Oh I misread the problem, it meets O1 and not O2 at G

Delray wrote: $A$, $F$ and $H$ aren't collinear, so how do we know that this works? orl wrote: Assume $ H$ is an arbitrary point on line segment $ AF.$

Here's a synthetic sol. - 1. Note that $ACBF$ is a kite with $AF=AE$ etc... 2. Claim : $\triangle AHE \sim \triangle ABD$ proof - $\angle HAE=\angle BAD$ ; Now let $H_1$ be the second intersection point of $EH$ and $O_2$ . Note that the map $H$ to $H_1$ is projective and $H_1$ to $H$ is projective , so $H$ to $G$ is a projective map , choose $H$ = $F , A$ and mid point of $FA$ and we get $\angle AEH = \angle ADB$ , so this is true for all positions of $H$ , so the claim is proved ! Now , $AH/HB$ = $AB/AD$ $AH = AB.AE/AD = AC^2/AD$ , since $AC = AF$ we are done !!! NOTE : To avoid diagram issues , I use projective maps !

Porky623 wrote: Hmm don't these solutions assume BECG is cyclic, which clearly isn't true because BEC are on O1 and G isn't? It's actually typed wrong on the AoPS wiki, but it is uneditable. I spent so much time trying to figure out what I did wrong, but really it was a flaw in the problem statement here: https://artofproblemsolving.com/community/c3667_2002_china_girls_math_olympiad - could somebody please fix this?

blacksheep2003 wrote: Porky623 wrote: Hmm don't these solutions assume BECG is cyclic, which clearly isn't true because BEC are on O1 and G isn't? It's actually typed wrong on the AoPS wiki, but it is uneditable. I spent so much time trying to figure out what I did wrong, but really it was a flaw in the problem statement here: https://artofproblemsolving.com/community/c3667_2002_china_girls_math_olympiad - could somebody please fix this? I think it's a typo in EGMO as well. I am not sure if it is fixed in some updated version but the version I've bought it isn't. I just found out and I'm fuming. I spent ~2 hours redrawing the diagram and be like "there's no way my compass and ruler diagrams are this far off I'm not this bad" before looking on the contests page and realizing it was a typo. Took like 5 minutes to do after I found out, what a waste of time...

Why they keep confusing people with such ugly formulation. I just can help it but correct. Quote: In $\triangle ABC$ with $\angle C=90^\circ$ altitude $CE$ meet $(ABC)$ at $F$. Point $H$ is on segment $AF$. $HE$ meet $(BCE)$ at $G$. $BG$ meet $AC$ at $D$. Prove that $FD \parallel CH$. The same in Taiwan 2016 TST1 Q1-2

Dear Mathlinkers, a very simple proof is possible... Sincerely Jean-Louis

AopsUser101 wrote:

instead of $\angle FBC = \angle GEC = \angle FEH = b$ do you mean $\angle GBC = \angle GEC = \angle FEH = b$

By angle chase we easily see that $AF=AC$, $BF=BC$ and that $FE=EC$ We denote with $x=\angle BCE$ and with $y=\angle FEH$. Then we have that: $$\frac{HF}{\sin y}=\frac{EF}{\sin 90-y+x}=\frac{CE}{\sin 90-y+x}$$ We have that $CE$ can be expressed in terms of $BC$, this means that $\frac{CE}{\sin 90-x}=BC$, but $BC$ can be expressed in terms of $CD$, $BC=\frac{\cos y}{\sin y}CD$ This implies that $HF=\frac{\cos y \cos x}{\sin 90-y+x}$. Similarily by noticing that $CF=2CE$ we have that $AF=\frac{\cos y \cos x}{\sin x \sin y}DC$ Thus our end goal is to prove that $(AF-HF)DC=AF.HF$, since that would imply $\frac{AH}{HF} = \frac{AC}{DC}$ Plugging in all of the information we have we get that $\cos y-x = \sin x \sin y + \cos x \cos y$ which is an identity this implies $\frac{AH}{HF} = \frac{AC}{DC}$

By the tangent-chord theorem, we have $\angle ACE=\angle CBE\coloneq 1$. Since $B\in\Gamma_2$, $\angle CFA=\angle 1$. Therefore, $\triangle ACF$ is isoceles. Moreover, it is easy to see that $\angle ECB=\angle FCB=\angle FAB\coloneq \angle 2$. Notice that $\angle BEC=90^\circ$ since $BC$ is diameter of $\Gamma_1$, and that $BA$ is diameter in $\Gamma_2$. Since $CF$ is a chord, $AB$ bisects $CF$, implying $FE=CE$. From this it also easily follows that $\angle EAC=\angle 2$. Define $\angle AEH\coloneq \angle 4$ We now observe $\angle HEF=\angle GEC=\angle GBC\coloneq \angle 3$. It follows $\angle CDB=\angle 4$. We now use the law of sines a couple of times. We have $$\frac{HE}{\sin(\angle 1)}=\frac{FH}{\sin(\angle 3)}$$and $$\frac{HE}{\sin(\angle 2)}=\frac{AH}{\sin(\angle 4)}$$in $\triangle HEF$ and $\triangle HEA$, respectively. This gives $$\frac{AH}{FH}=\frac{\frac{HE\sin(\angle 4)}{\sin(\angle 2)}}{\frac{HE\sin(\angle 3)}{\sin(\angle 1)}}=\frac{\sin(\angle 4)\sin(\angle 1)}{\sin(\angle 3)\sin(\angle 2)}$$In $\triangle BCD$, we have $$\frac{CD}{\sin(\angle 3)}=\frac{BC}{\sin(\angle 4)}$$Since $AEC$ is right-angled, we have $AC=\frac{AE}{\sin(\angle 1)}$. Combining gives $$\frac{AC}{CD}=\frac{AE\sin(\angle 4)}{BC \sin(\angle 1)\sin(\angle 3)}$$We thus want to show that $$\frac{\sin(\angle 4)\sin(\angle 1)}{\sin(\angle 3)\sin(\angle 2)}=\frac{AE\sin(\angle 4)}{BC\sin(\angle 3)\sin(\angle 1)} \Longleftrightarrow \frac{\sin(\angle 1)}{\sin(\angle 2)}=\frac{AE}{BC\sin(\angle 1)} $$But now notice that $\triangle CEB$ is right-angled, so $EC=BC\sin(\angle 1)$ Therfore, $$\frac{\sin(\angle 1)}{\sin(\angle 2)}=\frac{AE}{EC}\Longleftrightarrow \frac{AE}{\sin(\angle 1)}=\frac{EC}{\sin(\angle 2)}$$which is just the law of sines in $\triangle AEC$

Image:2002 CGMO-4.png We want to show that $$\frac{AH}{AF}=\frac{AC}{AD}$$but $AF^2=AC^2=AB \cdot AE$ so in fact it suffices to prove that $$AH \cdot AD=AB \cdot AE$$which is true if $$ \Delta AHE \sim \Delta ABD$$This follows readily from angle chasing: $ \angle HAE = \angle BAD$ is obvious, and $$\angle AEH = \angle BEG = \angle BCG = \angle ADB$$where the last equality is true because $BC \perp AD$ and $CG \perp DB$.

Dear Mathlinkers, here p. 20 Sincerely Jean-Louis

OMG, I was redrawing and redrawing but couldn't find the sol but reading above, the problem statement was wrong :skull: :skull:

illogical_21 wrote: \begin{align*} &=\frac{CE\cdot BG}{BE\cdot CG}\\ &=(EG,BC)\\ &\overset{B}{=}(DA,CP_\infty)\\ &=\frac{AC}{CD}, \end{align*} What do you mean by $(EG,BC)$ and how do you get that?

The main idea of this problem is to understand how $H$ relate to the points defined from it. Claim: $CGHF$ is cyclic. We have $$\angle CGH=\angle CBE=\angle CFH.$$ We will let $\theta=\angle HEF$ be our "moving variable". Note that $$CD=BC\tan\theta$$and from $CGHF$ cyclic we have $$HF=CG\cdot\frac{EH}{CE}=BC\sin\theta\cdot\frac{EH}{CE}.$$Thus, $$\frac{CD}{HF}=\frac{CE}{EH\cos\theta},$$so it suffices to show that $$\frac{CE}{EH\cos\theta}=\frac{AC}{AH}.$$Since we want ratios to share endpoints, rearrange this to $$\frac{CE}{AC}=\frac{EH\cos\theta}{AH}.$$ By Law of Sines on $\triangle AEH$, we have $$\frac{EH}{AH}=\frac{\sin\angle BAF}{\cos\theta},$$so $$\frac{EH\cos\theta}{AH}=\sin\angle BAF.$$Finally, since $$\sin\angle BAF=\sin\angle BAC=\frac{CE}{AC},$$we are done. remark: lol, missed the fact that $\triangle AHE\sim\triangle ABD$ just works. Lesson here: if there are two lengths that are not well understood, similar triangles can be used to reduce their ratio to a different ratio that is better understood.

Proving that \[\frac{AH}{AF} = \frac{AC}{AD},\] suffices. Notice that \[\angle AFC = \angle ABC = 90^\circ - \angle BAC = \angle ACE,\] which implies that $\triangle ACF$ is isosceles such that $AC=AF$. Hence, \[AC \cdot AF = AC^2 = AE \cdot AB\] upon which we need to prove that \[\frac{AB}{AD} = \frac{AH}{AE} \iff \triangle AHE \sim \triangle ABD.\] It is easy to see that $\angle HAE = \angle DAB$. Then, we note that \[\angle FAB = \angle FCB = \angle ECB = \angle EGB,\] so $\triangle AHE \sim \triangle GBE$, yielding $\angle AHE = \angle ABD$. Thus, the desired triangles are similar. $\square$

Note that $BCAF$ is a kite with perpendicular diagonals, from the right angles given in the problem statement. [asy][asy] pair B = dir(180); pair A = -B; pair C = dir(130); pair E = foot(C, A, B); pair F = B*B/C; pair D = 1.3*C-0.3*A; pair G = foot(C, D, B); pair H = extension(G, E, A, F); filldraw(unitcircle, invisible, blue); filldraw(circumcircle(B, E, C), invisible, red); draw(A--B, blue); draw(A--F--B--C--cycle, blue); draw(C--F, blue); draw(C--D, blue); draw(B--D, red); draw(G--H, red); draw(C--G, red); dot("$B$", B, dir(210)); dot("$A$", A, dir(A)); dot("$C$", C, dir(100)); dot("$E$", E, dir(235)); dot("$F$", F, dir(F)); dot("$D$", D, dir(D)); dot("$G$", G, dir(G)); dot("$H$", H, dir(290)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ B 210 = dir 180 A = -B C 100 = dir 130 E 235 = foot C A B F = B*B/C D = 1.3*C-0.3*A G = foot C D B H 290 = extension G E A F unitcircle / 0.1 yellow / blue circumcircle B E C / 0.1 yellow / red A--B / blue A--F--B--C--cycle / blue C--F / blue C--D / blue B--D / red G--H / red C--G / red */ [/asy][/asy] On the one hand, by law of sines repeatedly (and power of a point as $EB \cdot EA = EC \cdot EF$), we have \[ \frac{AH}{HF} = \frac{EA}{EF} \frac{\sin \angle AEH}{\sin \angle FEH} = \frac{EB}{EC} \frac{\sin \angle GEB}{\sin \angle CEG} = \frac{EB}{EC} \cdot \frac{GB}{GC}. \]This takes care of the points $F$ and $H$ which we can now erase. On the other hand, from similar right triangles $\triangle BCE \sim \triangle BCA$ and $\triangle BGC \sim \triangle BCD$, we get \begin{align*} AC &= EC \cdot \frac{CB}{EB} \\ DC &= GC \cdot \frac{CB}{GB}. \end{align*}Taking the quotient then completes the problem.

We claim that both the given ratios are equal to $$\dfrac{EC}{EB} \cdot \dfrac{BG}{GC}.$$Indeed, by the Law of Sines in $\Delta AHE$, we have $$AH = \dfrac{EH \sin HEA}{\sin EAH}.$$By the Law of Sines in $\Delta HEF$, we have $$HF = \dfrac{EH \sin FEH}{\sin EFH}$$Combined with angle chasing in cyclic quadrilaterals, it follows that \begin{align*} \dfrac{AH}{HF} &= \frac{\sin HEA \sin EFH}{\sin EAH \sin FEH} \\ &= \dfrac{\sin BEG \sin EBC}{\sin ECB \sin GEC} \\ &= \dfrac{EC}{EB} \cdot \dfrac{\sin BEG}{\sin GEC} \\ &= \dfrac{EC}{EB} \cdot \dfrac{\sin (90^{\circ} - GEC)}{\sin GEC} \\ &= \dfrac{EC}{EB} \cdot \cot GEC \\ &= \dfrac{EC}{EB} \cdot \cot GBC \\ &= \dfrac{EC}{EB} \cdot \dfrac{GB}{GC} \end{align*}as desired. Then, in right triangles $\Delta EAC$ and $\Delta GDC$, we have $$\begin{cases} AC = \frac{EC}{\sin EAC} \\ CD = \frac{GC}{\sin GDC} \end{cases}$$It follows that \begin{align*} \dfrac{AC}{CD} &= \dfrac{EC \sin GDC}{GC \sin EAC} \\ &= \dfrac{EC \sin BCG}{GC \sin ECB} \\ &= \dfrac{EC \cdot \dfrac{BG}{BC}}{GC \cdot \dfrac{BE}{BC}} \\ &= \dfrac{EC}{EB} \cdot \dfrac{BG}{GC} \end{align*}which leads to the desired condition. $\blacksquare$