Nagel

1. Nagel Point of a triangle

This is defined as the intersection point N of the lines joining the vertices A, B, C with the contact points A'', B'', C'' of the opposite
excircles of the triangle t = ABC. The Nagel point is the incenter of the anticomplementary triangle t' = A'B'C'.
The Incenter I the centroid G and the Nagel point N are on a line (called the Nagel Line of the triangle) and |GN| = 2*|IG| ([Bottema, p. 83]).

That these lines intersect at a point is shown by using the Ceva theorem (see Ceva.html ), taking into account the relations of BA''/A''C with the
sides of ABC. The assertion on the coincidence of the Nagel point with the incenter of the anticomplementary can be proved by considering
the relation between the trilinears (x,y,z) and (x',y',z') of the same point P w.r. to t and w.r. to t' (see Trilinears.html ).
Denoting by h_a, h_b, h_c the altitudes of t, we have:
                                                                                 x+x' = h_a,     y+y' = h_b,    z+z' = h_c.
This is a relation for the absolute trilinear coordinates of P. Taking x' = y' = z' = r' = inradius of t', we have
                                                                                           r'*(a+b+c) =4*S,
S being the area of t. This implies easily  that
                                                                               x = h_a - x' = 2(b+c-a)/(a*(a+b+c)).
Analogous formulas for y, z show that
                                                                              (x:y:z) = ( (b+c-a)/a  :  (c+a-b)/b  :  (a+b-c)/c  ),
which are the trilinears of the Nagel point.
Since t and t' are antihomothetic w.r. to their common centroid G, with ratio (-1/2) their incenters satisfy the given relation.

2. Other Nagel-like points

Below is drawn an extension of the figure defining the Nagel point. In this there are seen three additional Nagel-like points
{N₁, N₂, N₃} resulting as intersections of three cevians to the contact points with the tritangent circles of the triangle.
1) If point U is the contact of incircle with side AB then its antipode V is on the line CN_a.
2) The extension of CB₁ passes through the antipode of A₁.

First claim is a consequence of the homothety of the incircle to the excircle opposite to C. The homothety has center C and implies the
property that end-points of parallel radii of the two circles are aligned with C. This is the case with {C', V}.
Second claim holds for essentially the same reason. This time C is the (anti) homothety center of the two excircles with centers {A'', B''},
hence again end-points of anti-parallel radii are collinear with C. This is the case with {B₁, B₃}.
Analogous properties, of course, hold also for the other vertices of the triangle and the corresponding excircles and contacts.

3. A simple characterization

The following method ([Askwith, p. 13], rediscovered by [Hoehn]) gives another way to construct the Nagel point using only the incircle and
not the excircles. For this draw tangents to the incircle parallel to the sides. Then join the contact points of these parallels with the opposite
vertices. The three cevians thus created concure at the Nagel point.

The proof follows from [1] of the previous section, since the contact point with the parallel is the antipode V of U, and it was shown that
VC passes through the Nagel point.

4. Incidence problem

Show that the three Nagel-like points {N₁, N₂, N₃} joined to respective excenters {A'', B'', C''} define three concurring lines.
Compute the trilinears (barycentrics) of the point of concurrency W and show that it coincides with the De Longchamps point
(see De_Longchamps.html ).

Bibliography

[Askwith] E. H. Askwith, A course of pure Geometry Cambridge University Press, Cambridge 1903
[Bottema] O. Bottema Topics in Elementary Geometry Springer Verlag Heidelberg 2007
[Hoehn] Larry Hoehn A new characterization of the Nagel point Preprint, 2009

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1. Nagel Point of a triangle

2. Other Nagel-like points

3. A simple characterization

4. Incidence problem

See Also

Bibliography