Greetings!

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Dear Mathlinkers, any ideas? Sincerely Jean-Louis

Here is an intuitive way of doing it: We see that an inversion about $C$ of radius $\sqrt{CA.CB}$ composed with reflection in bisector of $\angle C$ maps this "problem" to the same problem in a similar triangle; the closer intersection is mapped to the farther one in the new figure. Using this similarity we get the isogonality. I don't know how to rigorous this. Any help?

Anant , it can be fixed if you slightly modify the argument. Consider the following tranaformation: Dilate wrt $C$ by factor $CA/CB$. Then reflect everything wrt angle bisector of $C$. Finally invert with power $CD^2$. It can be seen that after this transformation $\odot(AD),\odot(BE)$ are interchanged with one another. So $S$ has gone to $T$ and we are done.

Nice to see this, thanks Kapil! Here is a different approach. Consider inversion about $C$ with radius $\sqrt{CH.CF}$ where $F$ is the foot from $C$ to $AB$. This inversion maps the circle with diameter $AD$ to that with diameter $BE$. Now, their intersections are mapped to each other! (Since they are inverses) Since the intersection points are not fixed (check!) we get that $CS,CT$ are isogonal. Oops, turns out that both solutions are the same, except that yours is more clear (I need to dilate)

Well i dont think this inversion does it...plz check

Dear Mathlinkers, a proof without inversion is it possible? Sincerely Jean-Louis

Dear Mathlinkers, any ideas? Sincerely Jean-Louis

Is it true that $S$,$H$ and $T$ are collinear?

yes, consider the circle with diameter BC and finish with the thres chords theorem... Sincerely Jean-Louis

What is the three chord theorem????

I can roughly give an idea with barycentric coordinates. U can find the bar. Coordinate of S by finding the eqn. Of circle through A and D .now if that is (x,y,z) then prove that (c²x:a²y:b²z) satisfies the eqn. Of CT

Well, I didn't understand any of the inversion solutions and I think some may be probably wrong. That's my easy and clean solution (check out if it's fine): Consider an inversion centered at $C$ with radius $\sqrt{CE.CD}$ and reflect all trough the angle bisector of angle $ACB$. We will prove that $S$ and $T$ swap. Let $X$ and $Y$ be the projections of $D$ and $E$ onto $CA$ and $CB$. The circle $\omega_1$ pass trough $A$, $D$ and $X$ while $\omega_2$ pass trough $B$, $E$ and $Y$. By simple segment calculating, we get that: $CE.CD = CA.CY=CB.CX$ thus, our inversion and reflection makes: $A \rightarrow Y$ $B \rightarrow X$ $D \rightarrow E$ Thus, $\omega_1 \rightarrow \omega_2$ as we wanted.

Dear Mathlinkers, for the three chords theorem (Monge or d'Alembert) http://jl.ayme.pagesperso-orange.fr/Docs/Le%20theoreme%20des%20trois%20cordes.pdf Sincerely Jean-Louis

Der Mathlinkers, and without inversion? Sincerely Jean-Louis

RodSalgDomPort wrote: Is it true that $S$,$H$ and $T$ are collinear? Yes,it is.Just note that using Radical Axis Theorem on three circles $(AEBD)$, $\omega_{1}$ and $\omega_{2}$ (where $\omega_{1}$ and $\omega_{2}$ are circles with diameter $AD$ and $BE$,respectively) $AD$ and $BE$ concur on $ST$ and since this intersection point is $H\Rightarrow \overline{S-H-T}$.But how it can help?

Very nice problem ! . Here's a nice , elementary solution. Let $N,M$ be the respective midpoints of $AD,BE$ . Then it's clear that $\frac{CN}{CM}=\frac{ND}{ME}=\frac{NS}{MS}=\frac{NT}{MT}$ $(\bigstar)$ Also $\angle ACN=\angle BCM$ , so the internal angle bisector of $\angle NCM$ also bisects $\angle C$ , this yields \[\frac{NL}{LM}=\frac{CN}{NM}=(\bigstar)\]Now let the external angle bisector of $\angle NCM$ meet $NM$ at $K$ , it's clear that $\angle LCK=\frac{\pi}{2}\implies{(K,L;N,M)=-1}$ $(\clubsuit)$ $(\bigstar)\implies$ $SL$ bisects $\angle NSM$ $(\bigstar)\implies$ $TL$ bisects $\angle NTM$ $(\clubsuit)\implies$ $\angle LTK=\frac{\pi}{2}\implies{\{C,L,T,K\}}$ are concyclic on a circle with diameter $KL$. But it's well-known that $\{C,L,T,S\}$ lie on the $C$-Appolonian cirle of $\Delta NCM\implies{\{C,L,T,K,S\}}$ are concyclic. It's easy to see that $\Delta NSM\cong{\Delta NTM}\implies{CL}$ bisects $\angle SCT\Leftrightarrow{\angle SCL=\angle TCL}$. Hence , $\angle ACT=\angle ACL-\angle TCL=\angle BCL-\angle SCL=\angle BCS$ , as desired. $\blacksquare$

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