Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 732 details.
Saturday, February 25, 2012
Problem 732: Triangle, Altitude, Orthic Triangle, Orthocenter, Congruence, Parallelogram
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Let H be the ortho-center of ∆ABC
ReplyDeleteA₂C₁HB₁ 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
1) A₂B₁ ∥ A₁B₂ , B₂C₁ ∥ B₁C₂ and C₂A₁ ∥ C₁A₂
Delete2) <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