IMO Shortlist 2010 problem G5
Dodao/la:
arhiva23. lipnja 2013. Let
![ABCDE](/media/m/2/7/c/27c16cf5bf2e8ca59b13c61cf1562251.png)
be a convex pentagon such that
![AB = BC + AE,](/media/m/4/c/b/4cb7e4cdaaadb05be1f81652afb4f9df.png)
and
![\angle ABC = \angle CDE.](/media/m/0/d/6/0d622c3df5591de87c2d25765c444ac7.png)
Let
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
be the midpoint of
![CE,](/media/m/d/c/c/dcc8d0564fbfa598de2f20bfcb20cc40.png)
and let
![O](/media/m/9/6/0/9601b72f603fa5d15addab9937462949.png)
be the circumcenter of triangle
![BCD.](/media/m/a/e/d/aed6e0a9c7b3271bc8a26ed8dc66627c.png)
Given that
![\angle DMO = 90^{\circ},](/media/m/3/6/0/360f44362ecaba5d76e85f99504c9d2c.png)
prove that
![2 \angle BDA = \angle CDE.](/media/m/0/b/9/0b9dcd3748551b12ef1f9ca22cad5b71.png)
Proposed by Nazar Serdyuk, Ukraine
%V0
Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$
Proposed by Nazar Serdyuk, Ukraine
Izvor: Međunarodna matematička olimpijada, shortlist 2010