The Gergonne Point Theorem

The incircle touches each side at a point. The lines from each vertex to the opposite tangent point are concurrent at the Gergonne point, proven via Ceva's theorem and tangent length ratios.

Keywords: Gergonne point, incircle, tangent point, Ceva's theorem, concurrency

Prerequisites: cevas-theorem-proof, incenter-proof · Difficulty: intermediate

The Gergonne point is a triangle center defined by the incircle's tangent points. If the incircle of $\triangle ABC$ touches side $BC$ at $D$, side $CA$ at $E$, and side $AB$ at $F$, then the three lines $AD$, $BE$, $CF$ are concurrent.

Notation

Let $a = BC$, $b = CA$, $c = AB$ be the side lengths opposite the vertices $A$, $B$, $C$, and let

s = \frac{a + b + c}{2}
    

denote the semi-perimeter of the triangle.

Tangent Lengths

The two tangent segments from an external point to a circle have equal length. Applying this at each vertex of $\triangle ABC$ and solving the resulting three linear equations gives:

• From $B$: $BD = BF = s - b$
• From $C$: $CD = CE = s - c$
• From $A$: $AE = AF = s - a$

Applying Ceva's Theorem

\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB}
    = \frac{s-b}{s-c} \cdot \frac{s-c}{s-a} \cdot \frac{s-a}{s-b} = 1
    

Every factor cancels, so the product is $1$, and by Ceva's theorem the three cevians are concurrent at a point called the Gergonne point $Ge$.

The Proof

Step 1: Incircle and Tangent Points

Begin with triangle $ABC$. Construct the incircle, which touches each side at exactly one point: $D$ on $BC$, $E$ on $CA$, $F$ on $AB$.

Step 2: Cevians to Tangent Points

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

Step 3: The Gergonne Point Theorem

The Gergonne Point Theorem: the three cevians $AD$, $BE$, $CF$ from each vertex to the opposite incircle tangent point are concurrent. Their common point is the Gergonne point $Ge$.

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

Step 4: Tangent Ratios: Product $= 1$

Let $a = BC$, $b = CA$, $c = AB$ be the side lengths opposite each vertex, and let $s = (a+b+c)/2$ be the semi-perimeter.

The two tangent segments from each vertex to the incircle are equal, giving:

• $BD = s - b$, $DC = s - c$
• $CE = s - c$, $EA = s - a$
• $AF = s - a$, $FB = s - b$

\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB}
    = \frac{s-b}{s-c} \cdot \frac{s-c}{s-a} \cdot \frac{s-a}{s-b}
    = 1
    

By Ceva's theorem the cevians are concurrent.

Step 5: The Gergonne Point

By Ceva's theorem, since the product of the ratios is $1$, the three cevians $AD$, $BE$, $CF$ are concurrent. Their common point is the Gergonne point $Ge$.

Conclusion

The lines from each vertex to the point where the incircle touches the opposite side are concurrent at the Gergonne point $Ge$.

The proof is a direct application of Ceva's theorem: the tangent lengths from the incircle yield ratios whose product telescopes to $1$.