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Question In Space Geometry

  1. May 1, 2004 #1
    Hey everybody..
    I got this space geometry question that had me stumped.. Please help me out.. Here goes:

    ABCD is a parallelogram. The plane π was drawn passing through the diagonal AC.
    If DE was perpendicular to π, BF was perpendicular to π.
    Prove that BF = DE

    please see the attached drawing

    Attached Files:

  2. jcsd
  3. May 1, 2004 #2
    are E and F on π?
  4. May 1, 2004 #3
    yes, I guess they are.. since ED and BF are both perpendicular on π :confused: any help??
  5. May 1, 2004 #4


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    Well, what have you tried?
  6. May 2, 2004 #5
    Hey I think I got it!! :eek:
    First draw the diagonal BD

    DE and BF are both perpendicular to π
    so DE and BF are parallel

    This means that the angles BDE and FBD are equal.

    And since ABCD is a parellogram, then AB and CD are parrallel.
    so the angles BDC and DBA are equal

    We then conclude that the angles CDE ad ABF are equal
    (because we already proved that the bisects make equal angles)

    in the triangles DCE and ABF
    angle E=F=90 degrees
    D = B proven
    C = A 180-both angles

    so the triangles are congruent

    this means: ED/BF = CE/AF = CD/AB

    and since ABCD is a parallelogram, then CD = AB

    so DE/BF = CD/AB = 1/1

    that concludes to DE = BF :cool:

    Am I right here guys? :confused:
    Sorry for the crappy solution but I know all the theorems in arab so it's hard for me to write in english cuz I dont knwo the exact terms. :biggrin:

    Attached Files:

    Last edited: May 2, 2004
  7. May 6, 2004 #6
    :cool: I guess no newz is good news.. finally I can get this question off my back :rolleyes:
  8. May 6, 2004 #7


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    I'm not convinced that you've got it right. It's not clear how you show <CDE = <ABF. Sorry to put the question back on your back after all this time.
  9. May 6, 2004 #8


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    Try to look at the triangles DEM and BFM where M is the midpoint of AC.
    You've shown that <BDE = <DBF.
    And since M lies on AC, EM and FM are lines in the plane. Hence, <DEM =<BFM = 90.
    Finally, DM = BM = 1/2(BD) since the diagonals intersect at their midpoints.
    Thus, the triangles DEM and BFM are congruent; so DE = BF.

    'Congruent' means identical in shape and size, while the term 'similar' is used for objects that are identical in shape. So, when 2 triangles have the same angles, they are similar, but if they also have an equal side, they become congruent.

    Hope this helps.
  10. May 7, 2004 #9


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    I think this fills in the gap to first show that triangles ABF and CDE are similar.

    Since AB || CD, and is each cut by AC, then <BAC = <DCA.
    (Assuming that F and E are on AC, then <BAF=<DCE).

    As noted by xenogizmo, <E=<F.

    So, <ABF=<CDE.

    Thus, triangle ABF is similar to triangle CDE.
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