Sunday, May 16, 2010

Problem 455: Rhombus, Inscribed Circle, Angle, Chord, 45 Degrees

Geometry Problem
Click the figure below to see the complete problem 455 about Rhombus, Inscribed Circle, Angle, Chord, 45 Degrees.

Problem 455: Rhombus, Inscribed Circle, Angle, Chord, 45 Degrees.
See also:
Complete Problem 455

Level: High School, SAT Prep, College geometry

5 comments:

  1. It is enough to prove that E,O and G lie in the same line. Indeed, if we joing E with O and O with G we take:
    BOE=90-EOB and COG=90-OCG and BOC=90 because the diagons of a rhombus intersect creating an angle of 90 degrees.
    We sum and take EOC=270-(OCG+EOB)=180
    x=EH+JG/2=90/2=45

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  2. http://s004.radikal.ru/i208/1005/ef/0ef9e42743d2.gif

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  3. arc HFJ = 90° =>
    arc EJ + arc JG = 90° ( EOG altitude through O )
    EOH + GOJ = 90° =>
    EGH + JEG = 45° => ( EGH = 1/2 EOH, JEG = 1/2 GOJ )
    EKG = 135° =>

    GKJ = 45°
    -------------------------------------------

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  4. EG est un diamètre
    les triangles rectangles EOB et GOJ sont congruents
    théorème:
    la mesure d'un angle dont le sommet est à l'intérieur d'un cercle est égale à la demi-somme des mesures des arcs interceptés
    x=m(GKJ)=1/2(m(GOJ)+m(EOH))
    =1/2.90=45
    .-.
    =1/2.90=45

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  5. x=<GEJ+<HE=(<GOJ)/2+(<EOB)/2=90/2=45.

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