### Feuerbach's Theorem: a Proof

Feuerbach's Theorem states that the 9-point circle is tangent internally to the incircle of a triangle and externally to its excircles.

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Proof The proof that applies inversion to establish the tangency does this for pairs of circles. Following [Pedoe, pp. 9-10] and [Coolidge, p. 41], I'll do that for the excircles Cb and Cc lying opposite the vertices B and C, respectively. [Coxeter and Greitzer, p. 117-119] employ a similar method to the incircle and one of the excircles.

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Let S be the point of intersection of BC and CbCc. S is the external center of similitude of the two excircles, while A is their internal center of similitude. It follows that

 (1) ACb/ACc (= rb/rc) = SCb/SCc,

which means that points A and S divide the segment CbCc harmonically. The same holds for the projections P, B', C', S of the four points onto BC. Therefore,

 (2) PB'/PC' = SB'/SC'.

If L is the midpoint of the segment B'C', then (2) is equivalent to

 (3) LP·LS = LB'·LC' = LB'2 = LC'2.

This means that P and S are mapped on each other by inversion in the circle w centered at L of radius LB' (= LC'), i.e., the circle having B'C' as a diameter. Note that since BB' = (AB + BC + AC)/2 = CC', L is also the midpoint of the segment BC.

Let K be the midpoint of AB and M that of AC. The 9-point circle of ΔABC is the circumcircle of ΔKLM. Since it contains L, the inversion in circle w maps the 9-point circle onto a straight line through S. The 9-point circle is known to pass through the feet of the altitudes, P in particular. Therefore, the image of the circle under the inversion passes through S, the image of P.

To further identify this line, we invoke the angle preservation property of the inversion. Let XX' be the tangent to the 9-point circle at L. We have

 (4) ∠XLK = ∠KML = ∠ABC.

Because of the symmetry in the line of centers, the same angle is formed by AC and the line, other than BC, that is tangent to the two circles externally. In other words, XX' is parallel to that tangent. By the inversion at hand, KL is mapped onto itself. Also, KL||AC. The image of the 9-point circle must form the same angle with KL (or AC) as the circle itself. The second external tangent to the circles Cb and Cc fits the bill. It also passes through S. It is therefore exactly the image of the 9-point circle under the inversion. The circles Cb and Cc, being orthogonal to the circle of inversion, are anallagmatic, i.e., fixed under the inversion. So finally, since the image of the 9-point circle is tangent to the two circles, so is the 9-point circle itself.

As a variant, Hubert Shutrick suggested to take the tangent to the nine-point circle at K because L goes to infinity stretching the imagination of the reader. Since the angle between the tangent and KL is equal to ∠KML which is equal to the top angle in the diagram and since KL is parallel with AC, this is also the angle between KL produced and the line through S inverse to the nine-point circle.

### References

1. J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971
2. H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967
3. D. Pedoe, Circles: A Mathematical View, MAA, 1995 ### Inversion - Introduction ### Nine Point Circle 