Wednesday, February 23, 2011

Problem 582: Quadrilateral, Diagonals, Incenters, Metric Relations

Geometry Problem
Click the figure below to see the complete problem 582.

Problem 582: Quadrilateral, Diagonals, Incenters, Metric Relations.
Zoom at: Geometry Problem 582

Posted by Antonio Gutierrez at 2:43 PM
Labels: , , ,

2 comments:

  1. Peter TranFebruary 23, 2011 at 9:20 PM

    Connect OE,OF,OG,OH
    We have E,O,G collinear and F,O,H collinear ( angle bisectors of vertical angles)
    We also have EG perpendicular to FH at O ( angle bisectors of 2 supplement angles)
    apply Pythagoras therorem we have
    EF^2+HG^2=OE^2+OF^2+OH^2+OG^2=EH^2+FG^2

    Peter Tran

    ReplyDelete
  2. sourav mishraMarch 6, 2011 at 4:35 AM

    angle EOF is right angled as the bisectors of a linear pair enclose a right angle. so by applying Pythagoras theorem and associativity property of addition, we have EF^2 +GH^2 = (FO^2 + EO^2) + (HO^2 + GO^2) = (FO^2 + GO^2) + (EO^2 + HO^2) = FG^2 + EH^2.
    Q. E. D.

    ReplyDelete

Subscribe to: Post Comments (Atom)