Monday, August 26, 2013

Problem 915: Triangle, Intersecting Circles, Concyclic Points, Cyclic Quadrilateral

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

Click the figure below to see the complete problem 915.

Online Geometry Problem 915: Triangle, Intersecting Circles, Concyclic Points, Cyclic Quadrilateral.

Posted by Antonio Gutierrez at 9:02 PM
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5 comments:

  1. AjitAugust 26, 2013 at 10:01 PM

    Join D to F and extend it to D1. ///ly join G to F and extend to G1. Since ADFE is concyclic, /_EFD1 =/_A. ///ly /_GFD1=/B which implies that /_EFG1=/_C since A,B & C are internal angles of Tr. ABC. This, in turn, means that quad CEFG is concyclic.

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  2. UnknownAugust 27, 2013 at 5:41 AM

    This comment has been removed by the author.

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  3. nakaAugust 27, 2013 at 7:57 AM

    angle(EFG)
    = 360 - angle(DFE) - angle(DFG)
    = (180- angle(DFE)) + (180 - angle(DFG))
    = A + B
    = 180 - C
    = 180 - angle(GCE)
    q.e.d.

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  4. Jacob HA (EMK2000)August 27, 2013 at 7:59 AM

    ∠A + ∠DFE = 180°
    ∠B + ∠DFG = 180°

    ∠C + ∠EFG
    = (180°−∠A−∠B) + (360°−∠DFE−∠DFG)
    = 540° − (∠A + ∠DFE) − (∠B + ∠DFG)
    = 180°

    Thus, C,E,F,G concyclic.

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  5. AnonymousSeptember 1, 2013 at 10:53 AM

    Antonio,

    I would like to propose a new problem:

    D is located inside the triangle ABC, such that <DAC=<DAB=<DBA=10 degrees.
    <DBC is 30 degrees. Find <ACD.

    Respectfully,

    Erina-NJ

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