Saturday, September 12, 2009

Problem 354. Rhombus, Square, 45 degrees

Proposed Problem
Click the figure below to see the complete problem 354 about Rhombus, Square, 45 degrees.

Problem 354. Rhombus, Square, 45 degrees.
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Complete Problem 354
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 11:40 AM
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2 comments:

  1. Seja H o ponto de intersecção de CB e DE. O Ang(AGD) = Ang(CGH) = x. Assim, Ang(CDE) = Ang(DEC) = a, pois CE = CD. Ang(CHE) = 90° - a, pois o triângulo CEH é retângulo em C. O ang(DCA) = Ang(ACB) = b, pois a diagonal AC do losango (rombo) é também uma bissetriz do ângulo C. No triângulo CGH, temos que 90° - a = x + b, ou seja, x = 90° - (a + b), pelo teorema do ângulo externo. No triângulo CDG, temos que x = a + b, pelo mesmo teorema. Assim, x = 90° - x e, concluindo, temos que x = 45°(demonstrado).

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  2. join B with G and D.

    1) BG = GD (G equal distance to B,D)
    2) ang CBD = ang CDB ( BC=DC )

    => angCBG = angCDG (3)

    4) ang CFG = ang CDG ( FC = DC )

    from 3 and 4

    ang CBG = ang CFG

    but CF perpendicular to CB, so BG must be perpend to FG
    => BGD = 90
    => AGD = 45 ( CA bisector of C )

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