82. AA' , BN , CM concur in Gergonne point.

by Virgil Nicula, Aug 6, 2010, 10:00 PM

Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the side $BC$ at $A'$, and the line $AA'$ meets the incircle again at a point $P$. Let the lines $CP$ and $BP$ meet the incircle of triangle $ABC$ again at $N$ and $M$, respectively. Prove that the lines $AA'$, $BN$ and $CM$ are concurrent.
This difficult and very nice problem is an instructive aplication of the harmonical quadrilateral and the its properties. I suggest to see the characterization of the harmonical quadrilateral from the topic

Proof. I"ll use only the properties of the harmonical quadrilateral $\mathrm{(\ h.q.\ )}$ Define the tangent $XX$ in the point $X\in w$ to the incircle $w\ .$ Denote the intersection $R\in PP\cap BC$ and the points $B'\in AC$ , $C'\in AB$ which belong to the incircle $w$ of the triangle $ABC\ .$ Recall the relations $A'B=BC'$ and $A'C=CB'\ .$ $A\in C'C'\cap B'B'$ and $P\in AA'$ $\Longleftrightarrow$ $PB'A'C'\--\ \mathrm{h.q.}$ $\Longrightarrow$ $R\in B'C'$ and $\frac{PC'}{PB'}=\frac{A'C'}{A'B'}\ \ (1)\ .$ $\begin{array}{c} B\in C'C'\cap A'A'\ ,\ M\in PB\Longrightarrow PC'MA'\--\ \mathrm{h.q.}\Longrightarrow T\in A'C'\ ,\ PC'\cdot MA'=PA'\cdot MC'=\frac{1}{2}\cdot PM\cdot A'C'\\\ C\in BB'\cap A'A'\ ,\ N\in PC\Longrightarrow PB'NA'\--\ \mathrm{h.q.}\Longrightarrow S\in A'B'\ ,\ PB'\cdot NA'=PA'\cdot NB'=\frac{1}{2}\cdot PN\cdot A'B'\end{array}$ $\Longrightarrow$ $\frac{MP}{NP}=\frac{MC'}{A'C'}\cdot \frac{A'B'}{NB'}\ \ (2)\ .$ $\begin{array}{c}\triangle CB'N\sim\triangle CPB'\Longrightarrow NC\cdot PB'=B'C\cdot B'N\\\ \triangle BC'M\sim\triangle BPC'\Longrightarrow MB\cdot PC'=C'B\cdot C'M\end{array}$ $\Longrightarrow$ $ \frac{NC}{MB}=$ $\frac{B'C}{C'B}\cdot\frac{B'N}{PB'}\cdot\frac{PC'}{C'M}$ and $(1)$ $\Longrightarrow$ $\frac{NC}{MB}=\frac{B'C}{C'B}\cdot\frac{B'N}{C'M}\cdot\frac{A'C'}{A'B'}\ \ (3)\ .$ Observe that $R\in B'C'\Longrightarrow$ $\frac{RB}{RC}=\frac{A'B}{A'C}\ \ (3)\ .$ From the product of the relations $(2)$ , $(3)$ , $(4)$ obtain $\frac{RB}{RC}\cdot \frac{NC}{NP}\cdot\frac{MP}{MB}=1\ ,$ i.e. the points $R$ , $M$ , $N$ are collinearly $\Longrightarrow$ the lines $BN$ , $CM$ , $AA'$ are concurrently.
This post has been edited 5 times. Last edited by Virgil Nicula, Nov 23, 2015, 3:20 PM

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