The side $AB$ of a square $ABCD$ is the base of an isosceles triangle $ABE$ such that $AE=BE$ lying outside the square. Let $M$ be the midpoint of $AE$, $O$ be the intersection of $AC$ and $BD$. $K$ is the intersection of $OM$ and $ED$. Prove that $EK=KO$.
Problem
Source: Sharygin finals 8.5
Tags: geometry
TheDarkPrince
01.08.2018 21:11
Official Solution: Suppose $N$ is the midpoint of $BE$ and $L$ be the intersection of $ON$ and $EC$. Now its just noticing a parallelogram which by symmetry is a rhombus. My solution: 2 line coordinates.
Pluto1708
23.07.2019 16:27
Clearly $OM\parallel EC$ therefore $\angle KEO=\angle 90^{\circ}-\angle EDC=\angle OEC=\angle MOE=\angle KOE$. Q.E.D
Delta0001
23.07.2019 18:27
Wow, this is so easy for a Sharygin
jayme
17.02.2020 17:46
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20VII.pdf p. 46... Sincerely Jean-Louis