The Simson Line Theorem

A proof that the three perpendicular feet from a point on the circumcircle to the sides of a triangle are collinear, forming the Simson line.

Keywords: Simson line, circumcircle, perpendicular feet, collinear, cyclic quadrilateral

Prerequisites: cyclic-quadrilateral-proof · Difficulty: advanced

Let $P$ be a point on the circumcircle of triangle $ABC$. Drop perpendiculars from $P$ to sides $BC$, $CA$, $AB$, with feet $D$, $E$, $F$. Then $D$, $E$, $F$ are collinear — the line through them is the Simson line of $P$.

Strategy: show the rays $DF$ and $DE$ coincide by proving $\angle PDF=\angle PDE$. Two right angles at each foot make $PFBD$ and $PDCE$ cyclic; chasing inscribed angles through these circles and the circumcircle yields the equality.

Step 1: A Point on the Circumcircle

Start with triangle $ABC$ inscribed in a circle with center $O$. Place a fourth point $P$ on the circumcircle.

Step 2: Perpendicular Feet $D$, $E$, $F$

Drop perpendiculars from $P$ to each side of the triangle:

• $D$ is the foot on $BC$
• $E$ is the foot on $CA$
• $F$ is the foot on $AB$

Each perpendicular creates a $90°$ angle at its foot.

Step 3: The Simson Line

Connect the three feet $D$, $E$, $F$. They appear to lie on a single straight line — the Simson line.

We will prove this collinearity using cyclic quadrilateral properties.

Step 4: First Cyclic Quadrilateral $PFBD$

The proof relies on two auxiliary circles. Here is the first.

$\angle PFB=\angle PDB=90°$, so $P$, $F$, $B$, $D$ lie on the circle with diameter $PB$ (Thales' theorem: any point seeing $PB$ under a right angle is on this circle). Hence $PFBD$ is a cyclic quadrilateral.

Step 5: Second Cyclic Quadrilateral $PDCE$

Now the second circle.

Similarly, $\angle PDC=\angle PEC=90°$ places $P$, $E$, $D$, $C$ on the circle with diameter $PC$. So $PDCE$ is cyclic.

These two circles give the inscribed-angle equalities we'll chase in the next step.

Step 6: Inscribed Angles Chain to Collinearity

Goal: Show $\angle PDF=\angle PDE$, so rays $DF$ and $DE$ coincide at $D$, forcing $F$, $D$, $E$ onto a single line.

Two inscribed-angle equalities.

• In circle $PFBD$, chord $PF$ subtends equal angles at $D$ and $B$: $\angle PDF=\angle PBF=\angle ABP$ (since $F$ lies on $AB$).
• In circle $PDCE$, chord $PE$ subtends equal angles at $D$ and $C$: $\angle PDE=\angle PCE=\angle ACP$ (since $E$ lies on $CA$).

Bridge via the circumcircle. $P$, $A$, $B$, $C$ are concyclic, so inscribed angles on chord $AP$ give $\angle ABP=\angle ACP$.

Chaining: $\angle PDF=\angle ABP=\angle ACP=\angle PDE$. So $DF$ and $DE$ are the same ray from $D$, and $F$, $D$, $E$ are collinear.

Step 7: The Simson Line Theorem

We have proven that the perpendicular feet from any point $P$ on the circumcircle to the sides of triangle $ABC$ are collinear.

Converse: If the three perpendicular feet from $P$ are collinear, then $P$ must lie on the circumcircle. So the Simson line property is both necessary and sufficient for $P$ to be on the circumcircle.

For any point $P$ on the circumcircle of $△ABC$, the perpendicular feet from $P$ to the three sides are collinear. This line is the Simson line of $P$.

The converse also holds: if the three feet are collinear, then $P$ lies on the circumcircle. ∎