## Tuesday, November 24, 2009

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

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

Level: High School, SAT Prep, College geometry

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

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

3. 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

4. suggest to others

(i did not understand anything from third comment)

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

5. From E draw EH// BC ( H on GH)
Note 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

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

2. 6. Proved 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

7. Problem 394
Sulotion 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

8. Since GD bisects < BDC,
EP/PC = 1/sqrt2 and BG/GC = sqrt2

Dividing and since PC = CG, the result follows

Sumith Peiris
Moratuwa
Sri Lanka

9. AE=(sqrt2)*a/2
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