## Friday, February 24, 2012

### Problem 731: Cyclic quadrilateral, Triangle, Orthocenter, Altitude, Perpendicular, Parallelogram

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 731 details.

1. Note: Please draw a figure, as indicated.

Let O be the center of the circle.
Let M be the midpoint of AD.
Let G be the centroid of triangle BAD.
Let BM be the median of triangle BAD,
joining the vertex B to M.
Extend OG to E'such that GE' = 2OG
Consider triangles BE'G and GOM.
BG:GM = 2:1 (Property of median)
GE':OG = 2:1 (construction of E')
So BG:GM = GE':OG
Also BE' is parallel to OM
=> BE' is perpendicular to AD.
=> E' lies on the altitude from B (on AD)
By a similar argument E' lies on the altitude from A on the side BD (of triangle BAD.
So E' is the orthocenter of triangle BAD.
Hence E' coincides with E.
By proportionality of corresponding sides in the similar triangles BEG and OMG,
BE:OG = BG:GM = 2:1
Thus BE = 2 OM.
Noting that same O is the circumcenter and F is the orthocenter of triangle ACD, it follows by an exactly similar argument that CF = 2OM
Hence BE = CF.
Also BE is parallel to CF (each being perpendicular to AD)
Hence BEFC is a parallelogram.

2. http://img715.imageshack.us/img715/3812/problem731.png
BE and CF cut the circle at E’ and F’
Note that AD is the perpen. Bisector of EE’ and FF’ ( property of orthocenters)
And EFF’E’ and BCF’E’ are isosceles trapezoids
And angle(BE’F’)=angle(E’EF)=(E’BC) => BC//EF => BCFE is a parallelogram

3. Problem 731
Ιf O is the center of the circle ABCD and OM perpendicular AD .Τhen BE=//2.OM
CF=//2.OM. So BE=//CF therefore BCFE is parallelogram.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE