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

Labels:
altitude,
congruence,
orthic triangle,
orthocenter,
parallelogram,
triangle

Subscribe to:
Post Comments (Atom)

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