The circle $\Gamma$ and the line $\ell$ have no common points. Let $AB$ be the diameter of $\Gamma$ perpendicular to $\ell$, with $B$ closer to $\ell$ than $A$. An arbitrary point $C\not= A$, $B$ is chosen on $\Gamma$. The line $AC$ intersects $\ell$ at $D$. The line $DE$ is tangent to $\Gamma$ at $E$, with $B$ and $E$ on the same side of $AC$. Let $BE$ intersect $\ell$ at $F$, and let $AF$ intersect $\Gamma$ at $G\not= A$. Let $H$ be the reflection of $G$ in $AB$. Show that $F,C$, and $H$ are collinear.
Problem
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Tags: geometry, geometric transformation, reflection, power of a point, geometry unsolved
Luis González
10.11.2010 01:27
Let $T \equiv \ell \cap AB$ and $G' \equiv FC \cap \Gamma,$ different fom $C.$ By Pascal theorem for the degenerate cyclic hexagon $EEBACG',$ the intersections $D \equiv EE \cap AC,$ $F \equiv EB \cap CG'$ and $ T' \equiv AB \cap G'E$ are collinear $\Longrightarrow$ $T' \equiv AB \cap \ell,$ hence $T \equiv T'.$ Quadrilateral $FAET$ is cyclic on account of $\angle AEF=\angle ATF=90^{\circ}.$ Therefore $\angle AFT= \angle AEG'=\angle AGG'$ $\Longrightarrow$ $GG' \parallel FT$ $\Longrightarrow$ $G'$ is the reflection of $G$ across $AB,$ i.e. $H \equiv G'.$
thanhnam2902
10.11.2010 11:11
The same probolem at here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=358712
jayme
10.11.2010 14:11
Dear Mathlinkers, according my sources, this problem comes from Problème 5, German pre TST 2005 Sincerely Lean-Louis
djmathman
26.12.2013 19:22
Let $X=AB\cap \ell$, and let $O$ be the center of $\Gamma$. Since $DE$ is tangent to $\Gamma$, we have $\angle DEB=\angle BAE$, and since $\angle DXO=\angle DEO$ we have $DXEO$ cyclic, which implies $\angle AOE=\angle EDF$. Therefore $\triangle AOE\sim\triangle EDF$, and since $OA=OB$ we have $ED=DF$. Next, by Power of a Point, \[DC\cdot DA=DE^2=DF^2\implies \dfrac{DF}{DC}=\dfrac{DA}{DF}.\] Therefore $\triangle DCF\sim\triangle DFA$. Finally, since $GH\perp AB$, we have $GH\parallel\ell$, so $\angle DCF=\angle DFA=\angle HGA=\angle HCA$, whence $H,C,F$ are collinear.
