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 $XE = \frac{1}{2}AD$.
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 $\triangle MXN$ 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 $DF \parallel ME$ 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 $\angle FXD = \angle FXA + \angle AXD = \angle FXA + \angle XAF = 90$ and combined with the fact that $E$ is the midpoint of $FG$, we get that $\frac{AD}{2} = EF = EG = EX$.