Let $\Gamma$ be a circunference and $O$ its center. $AE$ is a diameter of $\Gamma$ and $B$ the midpoint of one of the arcs $AE$ of $\Gamma$. The point $D \ne E$ in on the segment $OE$. The point $C$ is such that the quadrilateral $ABCD$ is a parallelogram, with $AB$ parallel to $CD$ and $BC$ parallel to $AD$. The lines $EB$ and $CD$ meets at point $F$. The line $OF$ cuts the minor arc $EB$ of $\Gamma$ at $I$. Prove that the line $EI$ is the angle bissector of $\angle BEC$.
Problem
Source: http://oim2013.opm.org.pa/pdfs/examen_pt.pdf
Tags: geometry, parallelogram, geometry proposed
professordad
13.08.2014 21:09
Since $\overline{AE} \parallel \overline{BC}$ we have $\angle CBE = 45^{\circ}$. Also, $\angle BCD = \angle BAD = 45^{\circ}$. We have $\angle CDE = 45^{\circ}$ so $\overline{BE} \perp \overline{CD}$.
We see $BCED$ is cyclic. Also, $\angle BOD = \angle BFD = 90^{\circ}$ so $BODF$ is cyclic. So $\angle BEC = \angle BDC = \angle BOF = \angle BOI = 2\angle BEI$. The result follows.
rgreenanswers
13.08.2014 22:01
I know it seems really obvious but how do you prove that BCED is cyclic?
professordad
13.08.2014 22:32
$\angle CBE = \angle CDE = 45^{\circ}$
sayantanchakraborty
05.10.2014 21:20
Its again a basic problem,because I got it even after a heavy dinner. First of all it is easy to see that $\angle{BAE}=\angle{BEA}=45^{\circ}$.So an easy angle chasing will yeild that $BDEC$ is an isosceles trapezoid.Also note that since $DF \parallel AB \implies DF \perp BE$.$B$ is the midpoint of arc $AE$,so $\angle{BOD}=90^{\circ}$.Thus $BODF$ is cyclic.Now setting $\angle{BEI}=\theta$ we get that $\angle{DOF}=\angle{EOI}=90-2\theta$.Thus $\angle{BDF}=\angle{BOF}=2\theta$.Finally as $BDEC$ is cyclic we get $\angle{BDF}=\angle{BDC}=\angle{BEC}=2\theta$ so $\angle{BEI}=\angle{CEI}=\theta$ and $EI$ bisects $BEC$ as desired.