Friday, March 21, 2014

Geometry Problem 995: Quadrilateral, Perpendicular Diagonals, Perpendicular Bisector, Parallel

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view the complete problem 995.

Online Geometry Problem 995: Quadrilateral, Perpendicular Diagonals, Perpendicular Bisector, Parallel.

Posted by Antonio Gutierrez at 4:53 PM
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4 comments:

  1. Peter TranMarch 21, 2014 at 5:59 PM

    Since ∆AEC and ∆BFC are isosceles triangles
    So ∠BFD=180- 2. ∠ODA
    And ∠CEA=180-2. ∠CAE
    And ∠BFD+∠CEA= 360-2.( ∠ODA+∠CAE)= 180
    So BF//CE

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  2. APOSTOLIS MANOLOUDISSeptember 3, 2016 at 5:58 AM

    Problem 995
    Let <EAC=<ECA=x,<FBD=<FDBX+Y=y.But in triangle AOD apply <DAO+<ADO=90 or x+y=90. If BF intersects the AC at K then <BKO=90-<KBO=90-y=x=<AKF=<ACF. So
    BF//CE.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  3. Sumith PeirisJanuary 31, 2021 at 11:19 AM

    Let < NFB = p = < NFD = < CAE = < ACE

    So < CED = 2p = < BFP and so BF // CE

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. AnonymousFebruary 1, 2021 at 6:34 PM

    Let m(MEA)=@(Read as Alpha)=>m(NFE)=90-@
    AME congruent to CME (SAS AM=MC,m(AME)=m(EMC)=90)
    => m(MEC)=m(MEA)=@ and m(AEC)=2@ -------(1)
    Simiarly FNB congruent to FND
    => m(BFN)=m(NFD)=90-@=>m(AFB)=2@--------(2)
    From (1)&(2) BF//CE

    ReplyDelete

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