## Wednesday, July 28, 2010

### Problem 484: Square, Angle, 90 degrees, Triangle, Measurement, Proportion

Geometry Problem
Click the figure below to see the complete problem 484 about Square, Angle, 90 degrees, Triangle, Measurement, Proportion.

Complete Problem 484

Level: High School, SAT Prep, College geometry

1. Denote (XYZ) =angle XYZ
So polygon ABECD is cyclic with AC as a diameter of circumcircle of the polygon
In this circumcircle , arc BA=arc AD=arc DC= ¼ of full circle = 90
(BEA)=(AFD)=(DEC)= 45 ( These angles face 90 degrees arcs )
Consider triangle BEG and angle BEG, EF is internal angle bisector of ( BEG)
So FB/FG=EB/EG ( property of internal angle bisector)

Since EC perpen. To EF , EC will be external angle bisector of (BEG)
And CB/CG= EB/EG ( property of external angle bisector)

So FB/FG= CB/CG

Peter Tran

2. Problem 484
Is ABECD cyclic (<AEC=90=<ABC).Then <BEA=45=<AED=<DEC.So EG is internal bisector and the EB is external bisector of triangle FEC.Then FG/GC=EF/EC and EF/EC=BF/BC.
So FG/GC=BF/BC or BF/FG=BC/CG.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

3. Triangles BEG and CDG are similar so

BE/EG = CD/CG.....(1)
EF bisects < BEG hence

BE/EG = BF/FG...(2)

From (1) and (2) BC/CG = BF/FG

Sumith Peiris
Moratuwa
Sri Lanka