Saturday, February 25, 2012

Problem 732: Triangle, Altitude, Orthic Triangle, Orthocenter, Congruence, Parallelogram

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 732 details.

Online Geometry Problem 732: Triangle, Altitude, Orthic Triangle, Orthocenter, Congruence, Parallelogram.

2 comments:

  1. Let H be the ortho-center of ∆ABC
    A₂C₁HB₁ is a parallelogram.
    So diagonals A₂H and B₁C₁ bisect each other, say at X.
    B₂A₁HC₁ is a parallelogram.
    So diagonals B₂H and A₁C₁ bisect each other, say at Y.
    In ∆A₁B₁C₁,
    XY ∥ A₁B₁ and XY = ½ A₁B₁.
    In ∆HA₂B₂,
    XY ∥ A₂B₂ and XY = ½ A₂B₂
    Follows A₁B₁∥ A₂B₂ and A₁B₁= A₂B₂.
    Similarly we can prove that
    B₁C₁ = B₂C₂ and A₁C₁ = A₂C₂
    Hence ∆A₁B₁C₁ ≡ ∆A₂B₂C₂

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    Replies
    1. 1) A₂B₁ ∥ A₁B₂ , B₂C₁ ∥ B₁C₂ and C₂A₁ ∥ C₁A₂
      2) <A₂A₁C₂ = <A₁A₂C₁ and <A₁A₂C₂ = <A₂A₁C₁
      3) ∆A₂A₁C₂ = ∆A₁A₂C₁
      4) C₂A₁ = C₁A₂
      5) A₁C₂A₂C₁ is a parallelogram
      6) C₂A₂ = C₁A₁
      7) Similarly A₂B₂=A₁B₁ and B₂C₂=B₁C₁
      8) ∆A₁B₁C₁ ≡ ∆A₂B₂C₂

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