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Euler's Line Theorem


Date: 04/08/2001 at 14:13:41
From: Mahta
Subject: Euler's line theorem

Can you please help me to prove Euler's line theorem with analytical 
geometry - prove that the circumcenter, the orthocenter, and the 
centroid of any triangle lie on the same line?


Date: 04/09/2001 at 08:32:11
From: Doctor Floor
Subject: Re: Euler's line theorem

Hi, Mahta, thanks for writing.

Let us consider a triangle ABC. We choose coordinates in such a way 
that ABC lie on the unit circle, so the circumcenter O has coordinates 
(0,0). 

Let's say that A, B, and C have coordinates (a1,b1), (a2,b2), and 
(a3,b3), respectively. Since ABC lie on the unit circle, we have:

     (a1)^2 + (b1)^2 = (a2)^2 + (b2)^2 = (a3)^2 + (b3)^2 = 1

The centroid G has coordinates:

     ( (a1+a2+a3)/3 , (b1+b2+b3)/3 )

We will now show that the orthocenter H has coordinates:

     ( a1+a2+a3 , b1+b2+b3 )

If that is correct, then clearly O, H, and G are collinear, as 
desired. Their line is given by:

         b1+b2+b3
     y = -------- x
         a1+a2+a3

To prove the correctness of the coordinates for the orthocenter, we 
assume that H has the above coordinates, and show that H is indeed the 
orthocenter. First we compute the slopes of lines BC and AH:

     slope BC = (b2-b3)/(a2-a3)

     slope AH = (b2+b3)/(a2+a3)

Their product equals:

                            (b2-b3)(b2+b3)
     slope AH * slope BC =  --------------
                            (a2-a3)(a2+a3)

                            (b2)^2 - (b3)^2
                         =  ---------------
                            (a2)^2 - (a3)^2

now we use that (a2)^2 + (b2)^2 = (a3)^2 + (b3)^2 = 1:

                            1-(a2)^2 - (1-(a3)^2)
                         =  ---------------------
                               (a2)^2 - (a3)^2

                            (a3)^2 - (a2)^2
                         =  ---------------
                            (a2)^2 - (a3)^2

                         =  -1

This shows that AH and BC are perpendicular to each other. In exactly 
the same way we can show that BH and AC, as well as CH and AB, are 
perpendicular to each other, and thus that H with the above 
coordinates is indeed the orthocenter.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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