Monday, May 4, 2015

Geometry Problem 1118: Right Triangle, Angle Trisection, Concyclic Points, Cyclic Quadrilateral

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

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Online Math: Geometry Problem 1118: Right Triangle, Angle Trisection, Concyclic Points, Cyclic Quadrilateral.

Posted by Antonio Gutierrez at 2:59 PM
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3 comments:

  1. Peter TranMay 4, 2015 at 5:35 PM

    http://s12.postimg.org/a4sui5wi5/pro_1118.png
    Since B3C4 and CC4 bisect angles B1B3C and B3CB1 ( see problem 1116)
    So B1C4 will bisect angle B3B1C
    And ∠ (B3B1C4)= ½ . 60= 30 degrees
    And ∠ (C3B1C4)= ∠ ( C3B3C4)= 90 => B3, C3, B1, C4 are concyclic
    Similarly we also have B1A4 bisect angle AB1B3 and points A3, B3, A4 ,B1 are concyclic

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  2. Sumith PeirisAugust 7, 2015 at 3:00 AM

    Kindly refer my proofs for Problems 1114 thro to 1117. We have seen that < C3B1A4 = A4B1B2=< B2B1C4 = C4B1A3 = 30

    Hence < C3B1C4 is 90 and since < C3B3C4 = 90 C3B3C4B1 is cyclic


    Similarly A4B3A3B1 is cyclic

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Sumith PeirisAugust 7, 2015 at 3:03 AM

    Another problem arising from this-

    Prove B1B2A4C3 is cyclic and so is B1A3C4B2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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