The Nagel Point Theorem

The excircle opposite each vertex touches the opposite side at a point. The lines from each vertex to that tangent point are concurrent at the Nagel point, proven via Ceva's theorem.

Keywords: Nagel point, excircle, tangent point, Ceva's theorem, concurrency

Prerequisites: cevas-theorem-proof, excircles-intro · Difficulty: advanced

The Nagel point is a triangle center defined by the excircle tangent points. The excircle opposite vertex $A$ is tangent to side $BC$ at a point $D'$, and similarly for the other two excircles.

Excircle Tangent Lengths

For the excircle opposite $A$, the tangent lengths are:

• $BD' = s - c$ and $D'C = s - b$

(Note these are "reversed" compared to the incircle tangent lengths.)

Similarly:

• $CE' = s - a$ and $E'A = s - c$ (excircle opposite $B$)
• $AF' = s - b$ and $F'B = s - a$ (excircle opposite $C$)

Applying Ceva's Theorem

\frac{BD'}{D'C} \cdot \frac{CE'}{E'A} \cdot \frac{AF'}{F'B}
    = \frac{s-c}{s-b} \cdot \frac{s-a}{s-c} \cdot \frac{s-b}{s-a} = 1
    

The product telescopes to $1$, so by Ceva's theorem the cevians $AD'$, $BE'$, $CF'$ are concurrent at the Nagel point $Na$.

The Proof

Step 1: Excircles and Their Tangent Points

Every triangle has three excircles, one opposite each vertex. The excircle opposite $A$ is tangent to side $BC$ (and to the extensions of $AB$ and $CA$); its tangent point on $BC$ is $D'$.

Define $E'$ on $CA$ and $F'$ on $AB$ analogously, as the tangent points of the excircles opposite $B$ and $C$.

Step 2: Cevians to Tangent Points

Draw the cevian from each vertex to the excircle tangent point on the opposite side: $AD'$, $BE'$, $CF'$. We claim these three cevians are concurrent.

Step 3: The Nagel Point Theorem

The Nagel Point Theorem: the three cevians $AD'$, $BE'$, $CF'$ from each vertex to the opposite excircle tangent point are concurrent. Their common point is the Nagel point $Na$.

We'll prove it using Ceva's Theorem — the equal-tangent- length property of each excircle makes the Ceva product telescope to $1$.

Step 4: Tangent Lengths and Ceva's Product

Let $a = BC$, $b = CA$, $c = AB$ and $s = (a+b+c)/2$ be the semi-perimeter. The two tangent segments from each vertex to a given excircle are equal, which yields:

• $BD' = s - c$, $D'C = s - b$
• $CE' = s - a$, $E'A = s - c$
• $AF' = s - b$, $F'B = s - a$

(Note these are "reversed" compared to the incircle tangent lengths: the excircle opposite $A$ gives $BD' = s-c$ rather than $s-b$.)

Ceva's product becomes:

\frac{BD'}{D'C} \cdot \frac{CE'}{E'A} \cdot \frac{AF'}{F'B}
    = \frac{s-c}{s-b} \cdot \frac{s-a}{s-c} \cdot \frac{s-b}{s-a} = 1
    

Every factor cancels, so by Ceva's theorem the cevians are concurrent.

Step 5: The Nagel Point

By Ceva's theorem, the three cevians $AD'$, $BE'$, $CF'$ are concurrent. Their common intersection is the Nagel point $Na$.

The Nagel point is the isotomic conjugate of the incenter. It lies on the line from the incenter $I$ through the centroid $G$, with $G$ dividing $INa$ in the ratio $1:2$.

Conclusion

The lines from each vertex to the tangent point of the opposite excircle are concurrent at the Nagel point $Na$.

Like the Gergonne point, this follows from Ceva's theorem: the excircle tangent lengths yield a telescoping product equal to $1$.

The Nagel point is the isotomic conjugate of the incenter, and lies on the line from the incenter through the centroid (in ratio $1:2$).