Friday, October 5, 2018

Geometry Problem 1392: Two Cyclic Quadrilateral, Cyclic Octagon, Circle

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

Details: Click on the figure below.

Geometry Problem 1392: Two Cyclic Quadrilateral, Cyclic Octagon, Circle, Tutoring.

Posted by Antonio Gutierrez at 9:25 PM
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2 comments:

  1. c.t.e.oOctober 6, 2018 at 12:17 AM

    ang: MEF + PIJ = 180, MLK + GHP = 180 => AEML, PICH cyclic
    => EML + HPI = 180

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  2. Sumith PeirisOctober 6, 2018 at 11:28 AM

    Let the angles of Tr. FGB be a,b,x respectively

    Similarly for FGN be c,d,u
    For ELA be d,e,y
    For LEM be a,f,w
    For JKD be g,f, 180-x
    For JKQ be e,h,v
    For IHP be b,g, z and
    For HIC be c,h, 180 - y

    Write 8 sum of angles equations for the 8 Tr.s each adding to 180.

    Eliminate a,b,c,,d,e, f,g, h from these 8 equations

    We end up with u + v = w + z

    So each side must be = to 180 which is our desired result

    Sumith Peiris
    Moratuwa
    Sri Lanka

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