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Euler Line and Nagel Point


Date: 10/25/98 at 20:31:06
From: Wanwipa
Subject: Geometry

Can you provide more information on the Euler line and the Nagel point, 
including proofs?


Date: 11/09/98 at 11:37:20
From: Doctor Floor
Subject: Re: Geometry

Hi,

First the Euler line:

Let A' be the midpoint of BC, B' midpoint of AC and C' midpoint of AB.

               C
             /   \
            /      \               G: centroid.
           /   H     \             O: circumcenter.
          B'          A'           D: point on AB such that CD
         /       G       \            is perpendicular to AB.
        /         O        \       H: orthocenter.
       /                     \
      A--------D--C'-----------B

Take a careful look at triangles A'B'C' and ABC. You see that ABC and 
A'B'C' are similar, and that A'B'C' is half the size of ABC. Also AB 
and A'B' are parallel, AC and A'C' are parallel, and BC and B'C' are 
parallel, so that ABC and A'B'C' are sometimes called parallel 
triangles, or homothetic triangles.

Of course you see that AA', BB' and CC' meet in G. But then we know 
that ABC is the multiplication figure of A'B'C' with factor -2 over G. 
This means that if you take a point P of A'B'C', you measure the 
distance PG, multiply PG by two, and go that distance on the opposite 
side from G, you get the corresponding point of ABC.

Now A'O is perpendicular to BC - this way you construct O - but since 
BC and B'C' are parallel A'O is also perpendicular to B'C'. In this 
way we see that O is the orthocenter of A'B'C'. Aha, the circumcenter 
of ABC is the orthocenter of A'B'C'.

But now we must consider what happens when we multiply the orthocenter 
of A'B'C' over G with factor -2. It must go to the orthocenter H of 
ABC.

Conclusion: when you multiply O over G with factor -2, you get H. And 
thus the three points must be on one line.

Now the Nagel point:

The Nagel point is the point of concurrence of the three lines 
connecting the triangle vertices and the point where the excircle 
meets the opposite side.

Consider this line from the A-vertex:

                             /
                           C"
                         /
                       B            A'B"C" are the points where the
                     /  \           A-excircle meets the sides of ABC.
                   /     A'
                 /        \
               A---------- C----B"----------

Now we will multiply use the fact that the following:

Let X be a point, and Y and Z two points on a circle, such that XY and 
XZ are tangent to that circle. Then XY = XZ.

So AB" = AC", A'C = B"C, and BA' = BC". Knowing this it is easy to 
conclude that AB" = AC" = s (the semiperimeter, 0.5*(a+b+c)), 
BA' = s-c, and CA' = s-b. So we can conclude that BA':CA' = s-c : s-b.

Similar results can be found for the lines from B to B', and C to C':

   CB' : AB' = s-a : s-c
   AC' : BC' = s-b : s-a

In this way it is easy to see that (BA':CA')*(CB':AB')*(AC':BC') = 1 
and the concurrence of AA', BB', and CC' follows from Ceva's theorem 
(I suppose you know that one). And the Nagel Point is born.

I hope this helps! If you need more help on this or another question, 
write Dr. Math.

Best regards,

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

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