Feuerbach's Theorem (Metric Proof)

Feuerbach's theorem — the nine-point circle is tangent to the incircle — proved metrically by placing the circumcenter at the origin and showing the distance between the two centers equals the difference of their radii.

Keywords: Feuerbach's theorem, nine-point circle, incircle, tangent circles, Feuerbach point, barycentric coordinates

Prerequisites: ninepointcenter-proof, incenter-vector-proof, orthocenter-vector-proof, euler-oi-formula-proof · Difficulty: advanced

Feuerbach's Theorem (1822): the nine-point circle is tangent to the incircle — two circles from unrelated constructions, always touching at a single point, the Feuerbach point. With circumradius $R$ and inradius $r$ , the incircle (radius $r$ ) sits inside the nine-point circle (radius $R /2$ ), so the tangency is internal and the centers satisfy

∣ N I ∣ = \frac{R}{2} - r .

Strategy: prove this with vectors. Put the circumcenter $O$ at the origin, write $N$ and $I$ in coordinates from the two companion proofs, and compute $N - I$ . On screen we follow the centers and the segment $N I$ .

Step 1: The Incircle and Nine-Point Circle

We recall two circles of triangle $A B C$ , both established earlier: the incircle (center $I$ , radius $r$ ) and the nine-point circle (center $N$ , radius $R /2$ ).

Step 2: $∣ N I ∣ = R /2 - r$

Feuerbach's Theorem (1822). The nine-point circle is tangent to the incircle, at a single point $F$ — the Feuerbach point. The incircle lies inside, so the tangency is internal and the distance between the centers equals the difference of the radii:

∣ N I ∣ = \frac{R}{2} - r .

Step 3: Set Up Coordinates

Place the circumcenter $O$ at the origin, so $A$ , $B$ , $C$ are vectors of length $R$ . Two facts from the companion proofs:

the orthocenter vector formula $H = A + B + C$ , so $N = \frac{1}{2} (A + B + C)$ (the nine-point center is the midpoint of $O H$ );
the incenter coordinate formula $I = \frac{a A + b B + c C}{2 s}$ , where $2 s = a + b + c$ .

The one computational tool is the chord identity. The side $A B = c$ is the chord $A - B$ , so $∣ A - B ∣^{2} = c^{2}$ ; expanding the left side with $∣ A ∣ = ∣ B ∣ = R$ ,

∣ A - B ∣^{2} = ∣ A ∣^{2} - 2 A \cdot B + ∣ B ∣^{2} = 2 R^{2} - 2 A \cdot B = c^{2} .

Hence $A \cdot B = R^{2} - \frac{c ^{2}}{2}$ (and cyclically for the other pairs).

Step 4: Compute $∣ N I ∣$

Subtracting the two centers,

N - I = \frac{( s - a ) A + ( s - b ) B + ( s - c ) C}{2 s} .

Expanding $∣ N - I ∣^{2}$ with the chord identity gives

∣ N I ∣^{2} = \frac{R ^{2}}{4} - \frac{a ^{2} ( s - b ) ( s - c ) + b ^{2} ( s - c ) ( s - a ) + c ^{2} ( s - a ) ( s - b )}{4 s ^{2}} .

The bracket simplifies with $\sum (s - a) (s - b) = r^{2} + 4 R r$ and $(s - a) (s - b) (s - c) = r^{2} s$ (Heron's $r^{2} s^{2} = s \prod (s - a)$ ), collapsing to $4 s^{2} r (R - r)$ , so

∣ N I ∣^{2} = \frac{R ^{2}}{4} - r (R - r) = (\frac{R}{2} - r)^{2} .

Since $R \geq 2 r$ — Euler's inequality — the square root gives $∣ N I ∣ = \frac{R}{2} - r$ , exactly the difference of the two radii.

Step 5: Internally Tangent

Because the distance between the centers equals the difference of the radii, the incircle sits inside the nine-point circle and touches it at a single point — the Feuerbach point $F$ . That is Feuerbach's theorem.

Because $∣ N I ∣ = \frac{R}{2} - r$ is exactly the difference of the two radii, the incircle and nine-point circle are internally tangent at the Feuerbach point.

Extended theorem

Feuerbach proved more: the nine-point circle is tangent not only to the incircle but also to all three excircles, giving four points of tangency in all. The companion inversion proof gives a purely geometric view of why this happens. ∎