Thursday, December 13, 2012

Problem 831: Quadrilateral, Diagonal, Midpoint, Opposite Angles, Congruence, Parallelogram

Geometry Problem
Problem submitted by Jacob Ha, Keith Law, and more.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 831.

Online Geometry Problem 831: Quadrilateral, Diagonal, Midpoint, Opposite Angles, Congruence, Parallelogram

Posted by Antonio Gutierrez at 4:22 PM
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4 comments:

  1. Jacob HA (EMK2000)December 13, 2012 at 6:17 PM

    Solution of myself:

    Rotate 180° about O, then D→B, B→D.
    Let C→C', then C' is a point on line OA,
    with OC'=OC, so BCDC' is a parallelogram.

    Since ∠BC'D=∠BCD=∠BAD, C' coincide with A.
    Therefore, ABCD is a parallelogram.

    ******
    Solution submitted by Bean Ng:

    If OA=OC, then we are done.

    Otherwise, WLOG, suppose OA>OC,
    choose a point X on OA such that OX=OC.

    Then XBCD is a parallelogram.
    So ∠BXD=∠BCD=∠BAD.

    On the other hand, we have
    ∠BXD=∠BAD+∠ABX+∠ADX,
    thus ∠ABX=∠ADX=0.

    Hence, A=X (two points coincide)
    and so ABCD is parallelogram.

    ReplyDelete
  2. Jacob HA (EMK2000)December 13, 2012 at 8:00 PM

    I hope others will give some more elementary solution.
    i.e. not involve uniqueness argument.

    ReplyDelete
  3. AnonymousDecember 14, 2012 at 1:20 AM

    vectorAD+vectorDC=vectorAC
    vectorAB+vectorBC=vectorAC
    therefore, AD+DC=AB+BC
    vectorBA+vectorAD=vectorBD
    vectorBC+vectorCD=vectorBD
    therefore, BA+AD=BC+CD
    since ∠BAD=∠BCD
    thus, AD=BC and AB=DC
    and so ABCD is a parallelogram.

    ReplyDelete
  4. Jacob HA (EMK2000)December 15, 2012 at 5:44 AM

    Actually this problem is come from a variation of Problem 782 & 783.

    http://gogeometry.com/school-college/p782-triangle-orthocenter-circumcircle-diameter-parallelogram.htm

    http://gogeometry.com/school-college/p783-triangle-orthocenter-circumcircle-diameter-midpoint.htm

    ***
    The original problem is, as shown in 783:

    Let M be the mid-point of BC,
    produce HM to D, such that HM=MD,
    then BDCH is a parallelogram.

    Thus, ∠BDC=∠BHC. But ∠BHC=180°−∠A,
    so ∠BDC=180°−∠A, therefore ABDC is concyclic.
    Consequently, D lies on the circumcircle.

    ***
    Now is my variation:

    Let M be the mid-point of BC,
    produce HM to meet the circumcircle at D.

    I want to show that BDCH is a parallelogram.
    And it is somehow a converse of the above problem.

    Now I have BM=MC, and ∠BDC=180°−∠A=∠BHC,
    and hence Problem 831.

    I hope this can help anyone to give a elementary solution.

    ReplyDelete

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