See my observations here: http://forum.gil.ro/viewtopic.php?f=25&t=1171 Best regards, sunken rock

Dear Mathlinkers, for this result of de Longchamps and more, see http://perso.orange.fr/jl.ayme vol. 4 A new mixtilinear incircle adventure p. 21. Sincerely jean-Louis

This proof becomes easy when this point is known but have anyone a different solution?

The statement maybe false guy, I assumed that B' is on AC and C' is on AB since I don't know how B' on AB & C' on AC could make a question ^_^ Here with Cassey's theorem I got BC'/BA'=B'C/CA' , so when applying Ceva's theorem, it requires That AA' bisects $\angle BA'C$ to prove your problem, but That $AB=AC$ is not always true

Well, I have posted somewhere around here a problem using this fact, together with the proof, but it is quite long and boring, although fully synthetic; besides, I am not able to find it now. However, this link might help: link. Good luck, sunken rock

This is a result of monge's theorem.. We have that $A$ is the exsimilicenter of $ \Gamma_{a} $ and the incircle of $ABC$. Also $A'$ is the exsimilicenter of $\Gamma_{a}$ and $\Gamma$. Hence $AA'$ will pass through the exsimilicenter of $\Gamma$ and the incircle of $ABC$ which is a fixed point, call it $P$. Hence $AA',BB'$ and $CC'$ concurr at $P$.

There is a generalization of this result: Let $\Gamma$ and $\Gamma'$ two circles in the same plane. Points $X$, $Y$ and $Z$ are randomly set. Trace tangents $T_{X}$ and $T_{X}'$ from $X$ to $\Gamma'$ and let $\Gamma_{X}$ the circle tangent to $T_{X}$, $T_{X}'$ and $\Gamma$. Let $\Gamma\cap\Gamma_{X}=\{P_{X}\}$. Define in same fashion points $P_{Y}$ and $P_{Z}$. Prove that $\overline{XP_{X}}$, $\overline{YP_{Y}}$ and $\overline{ZP_{Z}}$ conccur. Demo: $X$ is the exsimilicenter of $\Gamma'$ and $\Gamma_{X}$, $P_{X}$ is the exsimilicenter of $\Gamma$ and $\Gamma_{X}$ so by Monge's theorem, the exsimilicenter of $\Gamma$ and $\Gamma'$ (let's call it $T$) lies on $\overline{XP_{X}}$. Repeat the argument for $\overline{YP_{Y}}$ and $\overline{ZP_{Z}}$, then you will see that these three lines contains $T$, son the concurrency is proven. In video (speaken into spanish, sorry)

Sorry for reviving, but I think this approach is way different. Let $A^*$ be the tangency point of the $A-$excircle and $BC$. We define $B^*$ and $C^*$ similarly. Lemma: $AA'$ and $AA^*$ are isogonal conjugates. Proof: Consider the inversion with center $A$ and radius $\sqrt{bc}$, which takes $A' \rightarrow A^*$. Back to the main problem. By Ceva, we know $AA^*, BB^*$ and $CC^*$ are concurrent. Therefore, by trig Ceva, we easily get $AA', BB'$ and $CC'$ are concurrent