Friday, July 24, 2015

Geometry Problem 1141: Inscribed Pentagon, Circle, Triangle, Congruence, Midpoint, Concurrency

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

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1141: Inscribed Pentagon, Circle, Triangle, Congruence, Midpoint, Concurrency.

Posted by Antonio Gutierrez at 9:28 AM
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1 comment:

  1. Pradyumna AgasheJuly 24, 2015 at 11:17 PM

    Consider Tr.ABC , lets assume Angle ACB = x and Angle BAC =y ,
    We get Angle ABC = 180 - (x+y)

    Tr. AFB and Tr. BMC are similar
    AF/FB = BM/CM
    Angle AFB = Angle BMC = 180 -x - y

    Now consider Tr.AFG and Tr. NMC
    AF/FG = MN/CM, angle AFG = angle NMC = x + y ,
    Angle AGF = Angle MCN , by AA similarity Tr.AFG is similar to Tr. APC
    Angle APC = Angle AFG = x + y
    Quad. ABCP is cyclic hence P lies on circle O.
    QED

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