Proposed Problem

Click the figure below to see the complete problem 394 about Square, 90 Degree Arc, Diagonal, Congruence.

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Complete Problem 394

Level: High School, SAT Prep, College geometry

## Tuesday, November 24, 2009

### Problem 394: Square, 90 Degree Arc, Diagonal, Congruence

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http://geometri-problemleri.blogspot.com/2009/11/problem-51-ve-cozumu.html

ReplyDeleteEF = a-a/V2 and FC=V2a-a where V=square root. So Another way to prove this w/o any construction would be: a=square side. EF/FC=a(1-1/V2)/a(V2-1)=1/V2 and ED/DC=(a/V2)/a =1/V2. Thus,ED/DC=EF/FC or DG bisects angle BDC. Therefore. BG/GC=DB/DC= aV2/a=V2/1 or BG/(BG+GC)=V2*/(V2+1)or rationalizing, BG =V2*a(V2-1) = 2a-aV2.

ReplyDeleteNow, EF=a-a/V2 or 2EF=2a - aV2. Hence, BG=2*EF.

Ajit

let us find the angles of angleBDG,BGD,DFE,EFD,DGC,GDC. then we can observe that DG is the angular bisector of angle BDC,and tr DEF, tr DGC are simalar triangles using these two concepts we can prove the required

ReplyDeletesuggest to others

ReplyDelete(i did not understand anything from third comment)

a easy way: use EF as middle line for any triangle

From E draw EH// BC ( H on GH)

ReplyDeleteNote that H is the midpoint of GD

And triangle EFH is isoceles ( EFH=FDA and FHE=FDA)

So EF=EH but EH=1/2. BG

So EF= ½. BG

Excellent approach to the solution. Mine was rather cumbersome trying to express EF and BG in terms of square side.

DeleteThank You

DeleteProved this by writing EF and BG in terms of a the side of the square and noting that Tr. CGF is isoceles but Peter's proof is far more easy and elegant

ReplyDeleteProblem 394

ReplyDeleteSulotion 1

Is <FDC=<FBC=<FAD/2=22.5=<FDB.If DF intersect AB at point P then <BPG=<FDC=22.5=<BDP so BP=BD=2BE. But triangle BPG and triangle EBF are

similar.Then BG/EF=BP/BE=2.Therefore PB=2BE.

Sulotion 2

Is <CGF=67.5=<AFD=<GFC so CG=CF .In triangle EDC DF is bisector so DE/DC=EF/FC (1)

and in triangle BDC DG is bisector so DC/DB=CG/BG (2). Μultiply by members of (1),(2)

σο ½=DE/DB=EF/GB or BG=2EF.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

Since GD bisects < BDC,

ReplyDeleteEP/PC = 1/sqrt2 and BG/GC = sqrt2

Dividing and since PC = CG, the result follows

Sumith Peiris

Moratuwa

Sri Lanka

AE=(sqrt2)*a/2

ReplyDeleteAF=AD=a (radii)

EF=AF-AE

=a-(sqrt2)*a/2

=(2-sqrt2)*a/2

In triangle EFD, <FDE=90

tan<EDF=EF/ED

=(2-sqrt2)/sqrt2

=sqrt2-1

<EDF=22.5

<GDC=45-22.5

=22.5

In triangle GDC

tan<GDC=GC/CD

GC=(sqrt2-1)*a

BG=BC-CG

=a-(sqrt2-1)*a

=(2-sqrt2)*a

=2EF