ISL 2002 G3

by Wolstenholme, Aug 4, 2014, 2:50 AM

The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$. Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$

Proof:

Since $ AE = AF $ we have that arcs $ \widehat{AE} = \widehat{AF} $ so $ CI $ bisects $ \angle{ECF} $. Therefore, since $ A $ is the midpoint of arc $ \widehat{EF} $ of circle $ S $, it suffices to show that $ AE = AF = AI $. Now since segments $ EF $ and $ AO $ are perpendicular and bisect one another we have that quadrilateral $ AEOF $ is a rhombus so $ AE = AF = OF = OD $ so it suffices to show that $ AI = OD $ so it suffices to show that $ AIOD $ is a parallelogram. But $ DO \perp AB $ by definition and $ AI \perp AB $ since $ BC $ is a diameter of $ S $ so $ AIOD $ is in fact a parallelogram and we are done.

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