Here are a few observations which I won't prove for now: (1) If (W) touches AB and AC in S, T respectively then I is the midpoint of ST. (2) PS and PT are the bisectors of angles <APB, <APC respectively. By using these we can prove what we want in the following manner: PA is the symmedian corresponding to P in triangle PST, because it's the intersection of the tangents through S and T to the circumcircle of PST. This means that <APS=<IPT=><APT+<IPT=<SPT and because <APT=<TPC=><IPT+<TPC=<SPT=><IPC=<SPT and in the same manner we show that <IPB=<SPT=<IPC=> I is on the bisector of angle <BPC, which is equivalent to what we wanted to show at the beginning.

Here's the proof of (2): Let S' and T' be the points where PS and PT cut the circumcircle of ABC (respectively). The circle (C) is obtained from (W) by a homothety of center P and ratio PT'/PT=PS'/PS, so the tangent in T' to (C) corresponds to the tangent to (W) in T (which is AC) by this homothety, so they're parallel, so T' is the midpoint of the arc AC. In the same manner we show that S' is the midpoint of the arc AB and the conclusion follows.

The "standard" proof for (1) uses Casey's thm and the transversal theorem (at least that's what it's called in Romanian). I'll post it in a little while if I fail to find another (nicer, non-computational) one.

Let A' be the foot of the bisector of <A in triangle ABC. From Casey's thm we get BS*AC+CT*AB=AS*BC (#). On the other hand, from the transversal thm we know that I is on ST iff (BS/SA)*A'C+(CT/TA)*A'B=(A'I/IA)*BC iff b*BS/SA+c*CT/TA=a which is true from (#) by division with AS=AT: AC*BS/AS+AB*CT/AT=BC (a, b, c are the sides of the triangle ABC, BC, CA, AB respectively). I guess it's not that long or ugly .

I can't find references to your Casey's theorem or transversal theorem... Could you please post the above two theorems formally?

Casey's theorem (there is a much more general form, using the notion of cycle instead of that of circle, but I won't post that, because I might make some mistakes): Let C be a circle and $C_i$ with i from 1 to 4 circles internally tangent to C, arranged around C in the increasing order of the indexes (1->2->3->4). Take $C_iC_j$ be the length of the external common tangent of $C_i$ and $C_j$. Then the analogous of Ptolemy's thm takes place: $C_1C_2\cdot C_3C_4+C_1C_4\cdot C_2C_3=C_1C_3\cdot C_2C_4$. The transversal theorem: ABC is a triangle, D is a point on BC, P is a point on AD, M and N are points on AB and AC respectively. Then the line MN passes through P iff DC*BM/MA+BD*CN/NA=BC*DP/PA, where all the 2-letter combinations represent oriented segments.

Sorry, but i don't understand how you can apply Casey's Theorom to this situation? please explain ...

You apply it for the circle W and the degenerate circles A, B, C. You can apply it for 2 circles and 2 points, 3 circles and a point ans other such cases. A point is nothing more than a degenerate circle.

Let us to present an alternative proof of the collinearity of the points $ P,\ I,\ D,$ based on the proof of the fact (1), mentioned by grobber in the above post #2#, which appeared in the topic Mixtilinear incircle of Leon. $ \bullet$ We denote the points $ B'\equiv (O)\cap BI,\ C'\equiv (O)\cap CI,$ where $ (O)$ is the circumcircle of the given triangle $ \bigtriangleup ABC$ and it is easy to show that $ B'C'\parallel ST,$ from $ \angle B'XC = \angle B'C'C + \angle ACC' = \frac {\angle B}{2} + \frac {\angle C}{2} = 90^{o} - \frac {\angle A}{2} = \angle ATS,$ where $ X\equiv AC\cap B'C'.$ So, because of $ IS = IT,$ from the trapezium $ B'C'ST,$ we have that the point $ Q\equiv B'C'\cap PI,$ is the midpoint of the segment $ B'C'$ and then, we conclude that $ OQ\perp B'C',$ where $ O$ is the circumcenter of $ \bigtriangleup ABC.$ $ \bullet$ It is easy to show that $ OQ = \frac {A'I}{2}$ $ ,(1)$ where $ A'\equiv (O)\cap AI,$ because of the incenter $ I$ of $ \bigtriangleup ABC,$ is the orthocenter of the triangle $ \bigtriangleup A'B'C'$ as well. So, from $ (1)$ and because of $ OQ\parallel IA',$ we conclude that the line segment $ IQ,$ passes through the point $ D\equiv (O)\cap A'O,$ as the symmetric point of $ A'$ with respect to $ O$ and then ( from $ OA'\perp BC$ ), as the midpoint of the arc $ BAC$ of $ (O).$ Hence, we conclude that the points $ P,\ I,\ D,$ are collinear and the proof is completed. Kostas Vittas.

Attachments:
t=5131.pdf (9kb)

Dear Mathlinkers, this result comes from Lauvernay (1892). See: http://perso.orange.fr/jl.ayme vol. 4 A new mixtilinear incircle adventure I, p. 17. Sincerely Jean-Louis

$\sqrt{bc}$-invert, which sends $T$ to the excentral touchpoint $D$, $I$ to the $A$-excenter $I_A$, and $T$ to the intersection point of the $A$-external bisector and $\overline{BC}$. It suffices to show that $A, D, I_A, T$ are concyclic now, but this is obvious.

Label the touch points to $AB$, $AC$, and $(ABC)$ as $K$, $L$, and $T$, and the midpoints of arcs $AB$ and $AC$ as $X$ and $Y$. Homothety tells us $TKX$ and $TLY$ collinear. Pascal on $BACXTY$ then tells us that $KIL$ collinear, and noting $\triangle AKI \cong \triangle ALI$ implies $IK = IL$. Thus $TI$ and $TA$ are a median and symmedian, respectively, of $\triangle TKL$, so $\angle KTA = \angle ITL = \frac C2$. As a result, it's clear that $TI$ will pass through the top point of $(ABC)$. $\blacksquare$