Saturday, September 6, 2014

Geometry Problem 1041: Quadrilateral, Perpendicular, Diagonals, Congruence, Metric Relations

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 1041.

Online Math: Geometry Problem 1041: Quadrilateral, Perpendicular, Diagonals, Congruence, Metric Relations.

3 comments:

  1. It is quite obvious that ABCD is an isosceles trapezium.
    That is, it is a cyclic quadrilateral.

    On the other hand, AC⊥BD and EF⊥AB.
    By Brahmagupta theorem, EF bisects CD.

    Since FG//HK, thus G is also the mid-point of CK.
    Hence, GK=3, and x=BK=7+3+3=13.

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  2. Draw CM //EG. M is on BE.
    Due to symmetric ∠(ABE)= ∠(ECD)
    But ∠(MCE)= ∠(MBF)
    So ∠(MCE)= ∠(ECD) => triangle DCM is isosceles and ED=EM
    We have ME/ED= CG/GK => GK=3 and x=13

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  3. Good work, Jacob and Peter

    It is possible to find AD from this which is = to 21/5 and so DC = 7/10 sqrt(208)

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