AoPS - Problem

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Tags: geometry, circumcircle, projective geometry, geometry unsolved

wangsacl

28.09.2010 08:14

Given an acute triangle $ABC$ with $AC>BC$ and the circumcenter of triangle $ABC$ is $O$. The altitude of triangle $ABC$ from $C$ intersects $AB$ and the circumcircle at $D$ and $E$, respectively. A line which passed through $O$ which is parallel to $AB$ intersects $AC$ at $F$. Show that the line $CO$, the line which passed through $F$ and perpendicular to $AC$, and the line which passed through $E$ and parallel with $DO$ are concurrent. Fajar Yuliawan, Bandung

Denote $R$ the intersection of $CO$ with the perpendicular line to $AC$ through $F.$ Then it is enough to show that $ER \parallel DO.$ Let the ray $CO$ cut $AB$ and the circumcircle $(O)$ at $P,Q,$ respectively. Since $QA \perp CA,$ it follows that $\triangle CFR$ and $\triangle CAQ$ are homothetic, and their cevians $FO$ and $AP$ are homologous $\Longrightarrow$ $\frac{_{CP}}{^{PQ}}=\frac{_{CO}}{^{OR}} \ (*).$ Because of $\angle ACQ=\angle BCE,$ it follows that $EQ \parallel AB.$ Hence, by Thales theorem $\frac{_{CP}}{^{PQ}}=\frac{_{CD}}{^{DE}},$ together with $(*)$ we obtain $\frac{_{CO}}{^{OR}}=\frac{_{CD}}{^{DE}}$ $\Longrightarrow$ $ER \parallel DO.$