How is the Simson line theorem proved?

The right angles at the feet put P, B, D, F on the circle with diameter PB and P, C, D, E on the circle with diameter PC, so the angles BDF and CDE equal the angles BPF and CPE at P. These are complements of equal angles at B and C (the exterior angle of the cyclic quadrilateral ABPC), so the two angles at D are equal vertical angles and F, D, E are collinear.

Is the converse of the Simson line theorem true?

Yes. If the feet of the perpendiculars from a point P to the three sides of a triangle are collinear, then P lies on the circumcircle of the triangle.

Simson Line

How to use

▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
Drag $P$ round the arc $BC$ , or $A$ , $B$ or $C$ round the circle; the readout under the canvas shows $\angle BDF$ , $\angle CDE$ and $\angle FDE$ . Learn more about why it holds
Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene

Proof?Drop the perpendiculars from P to the three sides. The right angles at D and F stand on PB, so P, B, D, F lie on a circle and ∠BDF = ∠BPF; likewise P, C, D, E lie on a circle and ∠CDE = ∠CPE. ABPC is cyclic, so its exterior angle at B equals the angle at C: ∠PBF = ∠PCE, and the complements ∠BPF and ∠CPE are equal. So ∠BDF = ∠CDE, and since B, D, C are on a line, F, D, E are on a line too. Drag P round the arc BC, or A, B, C.Simson line

DisplayLabelsMeasures

Drag P round the arc BC, or A, B or C round the circle

∠BDF = 15°,  ∠CDE = 15°,  ∠FDE = 180°

Step 1 / 6

Step explanations

1A point on the circumcircle and three feet

Given
P is a point of the circumcircle of triangle ABC, on the arc BC away from A. D, E, F are the feet of the perpendiculars from P to the lines BC, CA, AB (here F lies on AB extended beyond B). To prove: D, E and F lie on one line, the Simson line of P.

Drag P round the arc: the feet move together. Learn more about the statement

Getting Started with the Visualizer

The Statement

The First Circle

The Second Circle

The Cyclic Quadrilateral

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 7

Step through the proof, then drag

P

round the arc

BC

, or a vertex round the circle, and watch the three feet stay on one line: the angle

FDE

in the readout stays at

180°

. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.

Opening scene: the controls

Proof?Simson lineDisplayLabelsMeasuresDrag P round the arc BC, or A, B or C round the circle

∠BDF = 15°,  ∠CDE = 15°,  ∠FDE = 180°

← Back▶ PlayNext →ResetNormal ▾Step 1 / 6

Step explanations

1A point on the circumcircle and three feet

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1Proof — the theorem loaded in the player; the ? gives its idea in one line

2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc

3Draggable points — P slides round the arc BC, A, B and C slide round the circle; the hollow feet D, E, F follow

4Readout — the line under the figure: the angles BDF and CDE, and the angle FDE, live while you drag

5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure

6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen

7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it

Proof complete: step 6 of 6

Proof?Simson lineDisplayLabelsMeasuresDrag P round the arc BC, or A, B or C round the circle

∠BDF = 15°,  ∠CDE = 15°,  ∠FDE = 180°

← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6

Step explanations

1A point on the circumcircle and three feet

2P, B, D, F concyclic: ∠BDF = ∠BPF

Inscribed angles
The angles PDB and PFB are right angles standing on PB, so D and F lie on the circle with diameter PB (the converse of Thales). In that circle the angles at D and at P both stand on the arc BF: ∠BDF = ∠BPF = θ₁.

3P, C, D, E concyclic: ∠CDE = ∠CPE

Inscribed angles
In the same way the right angles PDC and PEC stand on PC, so D and E lie on the circle with diameter PC, and the angles at D and at P on the arc CE are equal: ∠CDE = ∠CPE = θ₂.

4∠PBF = ∠PCE

Cyclic quadrilateral
ABPC is a cyclic quadrilateral, so its opposite angles at B and C add up to 180°. The angle PBF is the exterior angle at B (F is on AB extended), so it equals the interior angle at C: ∠PBF = ∠ACP = ∠PCE = φ.

5θ₁ = 90° − φ = θ₂

Angle sum
The triangles PFB and PEC have right angles at F and E, so their other two angles add up to 90°: θ₁ = ∠BPF = 90° − φ and θ₂ = ∠CPE = 90° − φ. So θ₁ = θ₂, and therefore ∠BDF = ∠CDE.

6D, E, F are collinear

Vertical angles
B, D, C lie on one line, and F and E are on opposite sides of it. The rays DF and DE make equal angles with the opposite rays DB and DC, so they are vertical angles, which happens only when DF and DE are the two halves of one line: F, D, E are collinear. ∎

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1The Simson line — the dashed line through F, D and E

2The two small circles — on PB and on PC as diameters, each through two of the feet

3The angles at P — θ₁ and θ₂ between the perpendiculars and PB, PC

4Justification tags — Given, Inscribed angles (twice), Cyclic quadrilateral, Angle sum, Vertical angles

5Replay — at the last step Play becomes Replay and Next is off

The Statement

Take a triangle

ABC

and a point

P

on its circumcircle. Drop the perpendiculars from

P

to the three side lines, with feet

D

BC

E

CA

and

F

AB

. The theorem of the Simson line (found by William Wallace in 1797 and credited to Robert Simson) says that

D, \ E, \ F \text{ are collinear}

The line through them is the Simson line of

P

. For a point not on the circumcircle the three feet form a genuine triangle, the pedal triangle; the circumcircle is exactly where that triangle flattens into a line.

The player shows the usual case:

P

on the arc

BC

away from

A

, so that

D

falls on the side

BC

E

on the side

CA

, and

F

AB

extended beyond

B

. The readout shows the two angles at

D

that the proof compares, and the angle

FDE

, which is

180°

Opening scene, frozen

P on the circumcircle and the three perpendiculars, with feet D, E, F.

The First Circle

The angles

PDB

and

PFB

are right angles, and both stand on the segment

PB

. By the converse of Thales' theorem,

D

and

F

lie on the circle with diameter

PB

, so

P

B

D

F

are concyclic. In that circle the angles at

D

and at

P

both stand on the arc

BF

, and the inscribed angle theorem makes them equal:

\angle BDF = \angle BPF = \theta_1

The first circle, frozen

The circle on PB through D and F: ∠BDF = ∠BPF = θ₁.

The Second Circle

The same with

C

: the right angles

PDC

and

PEC

stand on

PC

, so

D

and

E

lie on the circle with diameter

PC

, and the inscribed angles on the arc

CE

give

\angle CDE = \angle CPE = \theta_2

The two angles at

D

that decide whether

F

D

E

are collinear have now been moved to

P

The second circle, frozen

The circle on PC through D and E: ∠CDE = ∠CPE = θ₂.

The Cyclic Quadrilateral

Now use the circumcircle itself.

A

B

P

C

lie on it in that order, so

ABPC

is a cyclic quadrilateral and its opposite angles at

B

and

C

add up to

180°

. Because

F

lies on

AB

extended beyond

B

, the angle

PBF

is the exterior angle at

B

, the supplement of

\angle ABP

; so it equals the interior angle at

C

\angle PBF = \angle ACP = \angle PCE = \varphi

(the last because

E

lies on

CA

The cyclic quadrilateral, frozen

The exterior angle at B equals the angle at C: both φ.

The Complements

The triangle

PFB

has a right angle at

F

, so its other two angles add up to

90°

(angle sum):

\theta_1 = \angle BPF = 90° - \varphi

. The triangle

PEC

has a right angle at

E

\theta_2 = \angle CPE = 90° - \varphi

. So

\theta_1 = \theta_2, \qquad \angle BDF = \angle CDE

The complements, frozen

θ₁ = 90° − φ = θ₂.

Why It Holds

The points

B

D

C

lie on one line, with

F

and

E

on opposite sides of it. The rays

DF

and

DE

make equal angles with the opposite rays

DB

and

DC

. Two angles like that are vertical angles exactly when their other arms are opposite rays too, so

DF

and

DE

are the two halves of one line:

F, \ D, \ E \text{ are collinear}

The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, inscribed angles in two auxiliary circles, the cyclic quadrilateral, the angle sum, and vertical angles.

The converse holds as well: if the feet of the perpendiculars from

P

are collinear,

P

lies on the circumcircle. For other positions of

P

the same chase works with the angles arranged differently; the player keeps to the one case, so the proof reads exactly as written.

The conclusion, frozen

The Simson line through F, D and E.

Dragging the Figure

Drag

P

round the arc

BC

: the three perpendiculars swing, the two auxiliary circles change size, the feet slide along the sides, and the Simson line turns, while the readout keeps

\angle BDF = \angle CDE

and

\angle FDE = 180°

. Drag

A

B

C

round the circle to change the triangle.

The player keeps to the case drawn:

P

on the arc

BC

away from

A

D

inside

BC

E

inside

CA

, and

F

AB

beyond

B

. Moving

P

far towards

C

would put

E

outside the side and the angles would have to be read differently, so such drags are refused, as are drags that crowd the points or leave the frame.

After dragging P, frozen

P slid round towards B: the feet moved, still on one line.

Consequences

• Diametrically opposite points. The Simson lines of two opposite ends of a diameter are perpendicular, and they meet on the nine-point circle.
• The orthocentre. The Simson line of

P

bisects the segment from

P

to the orthocentre

H

; its midpoint lies on the nine-point circle.
• Vertices. When

P

is a vertex, say

B

, two of the feet coincide with

B

and the Simson line is the altitude from

B

.
• The converse as a test. To show that four points lie on a circle, drop perpendiculars from one of them to the sides of the triangle of the other three and check that the feet are collinear.
• Angle between Simson lines. As

P

moves round the circle its Simson line turns by half the arc travelled, in the opposite direction, so the Simson lines envelop a curve with three cusps, the Steiner deltoid.