Geometry Problem
Click the figure below to see the complete problem 484 about Square, Angle, 90 degrees, Triangle, Measurement, Proportion.
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Complete Problem 484
Level: High School, SAT Prep, College geometry
Wednesday, July 28, 2010
Problem 484: Square, Angle, 90 degrees, Triangle, Measurement, Proportion
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Denote (XYZ) =angle XYZ
ReplyDeleteWe have ( ABC)=(AEC)=(ADC)=90
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
Problem 484
ReplyDeleteIs 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
Triangles BEG and CDG are similar so
ReplyDeleteBE/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
From similarity of ABF and CEF we get
ReplyDeleteBF/AB=EF/EC ........(1.)
From similarity of EFG and AED we get
EF/FG=AE/AD ........(2.)
From similarity of AEC and DCG we get
GC/CD=EC/AE ........(3.)
Multiplying equation 1 and 2 with AB=AD gives :
BF/FG=AE/EC
And from equation 3, with CD =BC we get:
BF/FG=CD/GC
BF/FG=BC/GC