Let $ABCD$ be a trapezoid with bases $AD$ and $BC$, and let $M$ be the midpoint of $CD$. The circumcircle of triangle $BCM$ meets $AC$ and $BD$ again at $E$ and $F$, with $E$ and $F$ distinct, and line $EF$ meets the circumcircle of triangle $AEM$ again at $P$. Prove that $CP$ is parallel to $BD$. Proposed by Ariel García
Problem
Source: Mexico National Olympiad Mock Exam 2018 Problem 1
Tags: geometry, trapezoid, circumcircle, parallel
plagueis
07.11.2018 06:12
Any solutions?
Prove that the reflection of $B$ in $M$ lies on the circumcirle of $\triangle AEM$.
TheDarkPrince
07.11.2018 07:58
Let $B'$ be the reflection of $B$ over $M$. We have \[\angle CEM = \angle CBM = \angle DB'M = \angle AB'M\]which gives $A,E,M,B'$ is cyclic. Now \[\angle MB'P=\angle MEP = \angle MEF = \angle MBF = \angle MB'C\]giving $B',C,P$ are collinear. As $B'C||BD$, $CP||BD$ and we are done.
MiltonMath12
07.11.2018 08:43
Other solution: Let $G$ be the intersection of $AD$ whit circuncircle of $\triangle AEM$. Note that $CG||BD$ because $\angle BFE=\angle BCA=\angle BCA$ but $\angle BCA=\angle CAD=\angle CPE$ therefore $CG||BD$. And $M$ is a midpoint of BG therefore $BCGD$ is a parallelogram and let P' be a intersection of $EF$ whit cincuncircle of $\triangle AEM$ and $\angle EP'C=\angle BFE$ but $EPGA$ is cyclic therefore $\angle EPC=\angle BFE$ and it follows that $P=P '$ or or lies same line with which we finished