Monday, December 5, 2011

Problem 698: Quadrilateral, Diagonal, Angle Bisector, Triangle, 120 Degrees

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

Click the figure below to see the complete problem 698.

Online Geometry Problem 698: Quadrilateral, Diagonal, Angle Bisector, Triangle, 120 Degrees.
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5 comments:

  1. AnonymousDecember 5, 2011 at 11:42 AM

    Let AB and DC intersect at P. Since m(APD)=60 and m(APD)+m(BEC)=180 PBEC is cyclic quadrilateral. PE is angle bisector of BAC. So, m(BEC)=m(EPB)=30

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  2. Peter TranDecember 5, 2011 at 5:01 PM

    http://img840.imageshack.us/img840/4838/problem698.png

    Extend AB and CD to F ( See picture)
    Note that angle(A)+ Angle (D)=120 and angle (AFD)=60
    Since AE and EC are angle bisectors of angles A and C so FE is angle bisector of angle ( AFD)
    Quadrilateral FBEC is cyclic ( F supplement to E) so x= angle (BFE)=30
    Peter Tran

    ReplyDelete
  3. PravinMarch 5, 2012 at 6:49 PM

    Problem 698 revisited:
    With F on AD let EF bisect angle AED.
    Clearly
    EA bisects angle BEF and
    ED bisects angle CEF
    So triangles AEB and AEF are congruent. So BE = EF.
    Similarly CE = EF.
    Hence BE = CE, triangle EBC is isosceles
    with vertical angle at E = 120 deg
    Hence each base angle = x = 30 deg

    ReplyDelete
    Replies
    1. AnonymousApril 2, 2012 at 11:58 PM

      wonderfully innovative solution

      Delete
  4. UnknownApril 25, 2012 at 9:30 AM

    The solution is uploaded to the following link:

    https://docs.google.com/open?id=0B6XXCq92fLJJbG9wSXBldmkweFE

    ReplyDelete

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