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.

 Problem 394: Square, 90 Degree Arc, Diagonal, Congruence.
See also:
Complete Problem 394

Level: High School, SAT Prep, College geometry

7 comments:

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

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  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

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  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

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  4. suggest to others

    (i did not understand anything from third comment)

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

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  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

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  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

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  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

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