Let $ABCD$ be a trapezoid with bases $AB$ and $CD$, inscribed in a circle of center $O$. Let $P$ be the intersection of the lines $BC$ and $AD$. A circle through $O$ and $P$ intersects the segments $BC$ and $AD$ at interior points $F$ and $G$, respectively. Show that $BF=DG$.
Problem
Source: Centroamerican Olympiad 2014, problem 2
Tags: geometry, trapezoid, geometry proposed
admin25
11.06.2014 02:54
Since OGPF is cyclic, $ \angle OGD=\angle OGP-180^\circ-\angle OFP=\angle OFB $. Also, $ OB=OC=OD=OA $ and $ AD=BC $, so $ \triangle OBC\cong\triangle ODA $, and $ \angle OBF=\angle OBC=\angle ODA=\angle ODG $. Therefore, $ \triangle OBF\sim\triangle ODG $. However, $ OB=OD $, so $ \triangle OBF\cong\triangle ODG $, and $ BF=DG $.
Jutaro
11.06.2014 03:24
I am the author of this problem. Your solution is the same as the one I had in mind.
juckter
11.06.2014 05:15
I'd think this would fit better as problem 1. Too easy really.
Bryan0224
28.10.2024 13:55
By obvious symmetry, $\measuredangle DGF=\measuredangle DPC=\measuredangle DPO=\measuredangle GPO=\measuredangle GFO$. Hence $OF=OG$. By Power of a point, $AG\cdot GD=R^2-DG^2=R^2-DG^2=R^2-DF^2=CF\cdot FB$. Since $ABCD$ is cyclic, $AD=BC$, hence $DG=BF$ or $FC$. It suffices to consider the case $DG=FC$. When $DG=FC$, $G$ is the reflection of $F$ over $OP$. Hence $\angle PGO=90^{\circ}$ since $PGOF$ is cyclic. Since $OA=OD$, $G$ is the midpoint of $AD$ and $F$ is the midpoint of $BC$. So $DG=FC=BF$. $\blacksquare$