Saturday, August 15, 2009

Problem 337. Isosceles Trapezoid, Angle bisector, Parallel, Concyclic points

Proposed Problem
Click the figure below to see the complete problem 337 about Isosceles Trapezoid inscribed in a circle, Angle bisector, Parallel, Concyclic points.

Problem 337. Isosceles Trapezoid, Angle bisector, Parallel, Concyclic points.
See more:
Complete Problem 337
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 8:50 AM
Labels: , , , , ,

2 comments:

  1. c .t . e. oJanuary 4, 2010 at 11:08 AM

    ang ADC + ang BCD = 180 (from ang of one side in trap)

    so need to prove ang EDH(C) + ang EGH = 180

    ang EDC = ang ADC + ADE (or ECD), (CE bisector)
    so for ang EDC have to add ECB ( BCD - ECD )
    but ECB = EBC ( equal bisector for equal ang )
    so ang EGH = ang EBC (are at the same side of L1 //BC)

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  2. Peter TranDecember 25, 2015 at 6:51 PM

    http://s17.postimg.org/p2m3e2le7/pro_337.png

    Connect DE , EG and EB
    Since E is the midpoint of arc AD and trapezoid ABCD is isosceles
    So EBC and EGF are isosceles triangles.
    And ∠ (GFE)= ∠(FGE)= ∠(BCE)= ∠(EBC)
    So B, G, E are collinear
    BCDE is cyclic => ∠(EDC) supplement to ∠(EBC)=> ∠(EDC) supplement to ∠(FGE)
    So EGHD is cyclic

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