Saturday, August 25, 2012

Problem 799: Parallelogram, Angle Bisector, Parallel, Triangle, Circumcircle, Circumcenter

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 799.

Online Geometry Problem 799: Parallelogram, Angle Bisector, Parallel, Triangle, Circumcircle, Circumcenter.

4 comments:

  1. http://img19.imageshack.us/img19/9383/problem799.png

    Draw angles and lines per attached sketch
    We have ∆ABE and ∆ECF are isosceles triangles
    QE=QC (Q is circumcenter of ∆CEF)
    AB=CD=BE
    ∠EQH=∠CQH=∠EFC= α
    ∠ECD=2 α ( external angle of ∆ ECF)
    ∠QCD= α +90 and ∠QEB=external angle of EQH= α +90
    So ∆BQE congruence to ∆DQC ( Case SAS) =>∠ BQD=2 α = ∠BCD
    So quadrilateral BDCQ is cyclic

    ReplyDelete
  2. Our target is to prove that B,Q,C,D concyclic.

    Let x be the angle FAD (=angle BAE = angle BEA = angle FEC = EFC) [from previous problem797]

    Since EFC is isos., QC is the perpendicular bisector of EF and angle QCE = 90-x. Then angle QCD = 90 + x.
    Now observe that
    QC = QE
    BE = BA = CD
    Angle QCD = 90 + x = 180 - (90+x) = 180 - (angle QEC) = angle QEB

    Hence triangle QEB is congruent to triangle QCD.

    Therefore angle QBC = angle QBE = angle QDC.
    By the converse of angle in the same segment, it is proved.

    ReplyDelete
  3. ∆'s ABE, CEF are isosceles and CQ ⊥ bisector of EF
    imply ∠QEB = Π - ∠QEC = Π - ∠QFC = Π - ∠QCF = ∠QCD
    So ∆QBE ≡ ∆QDC, ∠QBC = ∠QBE = ∠QDC
    Hence Q, B, D, C are concyclic

    ReplyDelete
  4. Also BD is tangential to circle BEQ

    ReplyDelete