Let ABCD be the parallelogram, such that angle at vertex A is acute. Perpendicular bisector of the segment AB intersects the segment CD in the point X. Let E be the intersection point of the diagonals of the parallelogram ABCD. Prove that XE=12AD.
Problem
Source: Polish Junior Math Olympiad 2nd Round 2020
Tags: geometry, parallelogram, perpendicular bisector
zuss77
22.10.2020 20:52
super easy, probably P1?
If M,N are midpoints of AB,CD then E - circumcenter of △MXN and EX=EN=AD/2
pggp
22.10.2020 21:01
It is 2nd stage of olympiad for 8th grade in Poland, so won't be hard.
franzliszt
22.10.2020 23:05
Fast sketch: Let the midpoint of AD be M. MDFE is cyclic and DF∥ME so MDFE is an isosceles trapezoid.
Attachments:

parmenides51
08.04.2021 20:25
posted and solved also here
ZJ42
08.04.2021 20:26
are we allowed to post answers to math olympiad problems?
franzliszt
08.04.2021 20:31
Yes you are lol
hakN
08.04.2021 21:05
Let F be the midpoint of AB and let G be the midpoint of CD. Obviously F,E,G are collinear. We have ∠FXD=∠FXA+∠AXD=∠FXA+∠XAF=90 and combined with the fact that E is the midpoint of FG, we get that AD2=EF=EG=EX.