Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/An%20unlikely%20concurrence.pdf Sincerely Jean-Louis

Let's define $P'=DE\cap MN$ and let's prove that $P'\equiv P$ showing that $P' \in AI$. Define $F=AI\cap BC$, $G=BP'\cap AC$ and $H=CP'\cap AB$. By Ceva Theorem, our claim is equivalent to $$\frac{BH}{HA}\cdot \frac{AG}{GC}\cdot \frac{CF}{FB}=1$$ So we will determine the value of each fraction. In particular we know that $$\frac{BH}{HA}=\frac{[CBH]}{[CHA]}=\frac{\frac{1}{2} BC\cdot CH\cdot \sin \angle BCH}{\frac{1}{2}CH\cdot AC\cdot \sin \angle HCA}=\frac{a}{b} \cdot \frac{\sin \angle BCH}{\sin \angle HCA}$$ By similar calculations on triangle $DCE$, which is isosceles, we obtain that $\frac{DP}{PE}=\frac{\sin \angle BCH}{\sin \angle HCA}$. Since $MP \parallel EC$, from Thales' Theorem we obtain that $\frac{DP}{PE}=\frac{DM}{MC}$. If we establish $MC$ to be positive, the following holds $$\frac{BH}{HA}=\frac{a}{b}\cdot \frac{DM}{MC}=\frac{2\cdot DM}{b}$$ Now, we are going to determine the value of $\frac{AG}{GC}$. From $MN\parallel AC$ and $\vert DM\vert =\vert PM\vert$ and from appropriate considerations on signs we get $$\frac{AG}{GC}=\frac{b-2\cdot DM}{2\cdot DM}$$ Applying the Internal Bisector Theorem we finally get $$\frac{CF}{FA}=\frac{b}{c}$$ So the following holds $$\frac{BH}{HA}\cdot \frac{AG}{GC}\cdot \frac{CF}{FB}=\frac{2\cdot DM}{b}\cdot \frac{b-2\cdot DM}{2\cdot DM}\cdot \frac{b}{c}=\frac{b-2\cdot DM}{c}$$ Little calculation shows that $DM$, taken with sign, equals to $\frac{b-c}{2}$, and so $$\frac{BH}{HA}\cdot \frac{AG}{GC}\cdot \frac{CF}{FB}=\frac{b-2\cdot DM}{c}=\frac{c}{c}=1$$Q.E.D