Triangle similarity

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Triangle similarity

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Source: Rioplatense Olympiad 2011, Level 3, Problem 2

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#1 h Aug 29, 2014, 8:56 PM • 1 Y

Y by Adventure10

Let $ABC$ an acute triangle and $H$ its orthocenter. Let $E$ and $F$ be the intersection of lines $BH$ and $CH$ with $AC$ and $AB$ respectively, and let $D$ be the intersection of lines $EF$ and $BC$ . Let $\Gamma_1$ be the circumcircle of $AEF$ , and $\Gamma_2$ the circumcircle of $BHC$ . The line $AD$ intersects $\Gamma_1$ at point $I \neq A$ . Let $J$ be the feet of the internal bisector of $\angle{BHC}$ and $M$ the midpoint of the arc $\stackrel{\frown}{BC}$ from $\Gamma_2$ that contains the point $H$ . The line $MJ$ intersects $\Gamma_2$ at point $N \neq M$ . Show that the triangles $EIF$ and $CNB$ are similar.

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#2 h Aug 30, 2014, 5:24 AM • 2 Y

Y by Adventure10, Mango247

$EF$ is radical axis of $\Gamma_1,$ $\odot(BCEF)$ and $BC$ is radical axis of $\odot(BCEF),$ $\odot(ABC)$ $\Longrightarrow$ $AD$ is radical axis of $\odot(ABC),\Gamma_1$ $\Longrightarrow$ $I \in \odot(ABC)$ $\Longrightarrow$ $I$ is center of the spiral similarity that swaps $\overline{CE}$ and $\overline{BF}$ $\Longrightarrow$ $IF:IE=BF:CE=HB:HC.$ But since $HJ$ and $NJM$ bisect $\angle BHC$ and $\angle BNC,$ then by angle bisector theorem $NB:NC=JB:JC=HB:HC$ $\Longrightarrow$ $NB:NC=IF:IE.$ Together with $\angle CNB=\angle EHC=\angle EIF,$ it follows that $\triangle EIF \sim \triangle CNB$ by SAS.

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