The Simson Line
by Emily Kennedy

Hopefully, your investigation yielded something like this:

The blue circles seem to lie on the circumcircle of the triangle.

So if the pedal point P lies on the circumcircle of ΔABC, it looks like we have a Simson Line.

Let's prove it!

Let P be any point on the circumcircle of ΔABC, and let D, E, and F be the vertices of the pedal triangle for pedal point P.

Draw in segments , , and , and label some angles as shown:

Look at quadrilateral ACPB.

A, C, P, and B all lie on a single circle, so
(β + γ) + δ = 180°
(1)
(Why? See 19.6.1 in this helpful PDF.)

Now look at quadrilateral AEPD.

= 90°, and = 90° (by construction, since ΔDEF is a pedal triangle).

And the sum of the interior angles of a quadrilateral is always 360°, so we have
δ + + (β + α) + = 360°
δ + 90° + (β + α) + 90° = 360°
δ + (β + α) = 180°
(2)

By (1) and (2), we have
(β + γ) + δ = 180° = δ + (β + α)
γ = α
(3)

Now I want to show that γ = σ

I claim that B, D, F, and P lie on a single circle.
How can we prove this?

First, take a look at this GSP animation.
This animation traces the locus of a point Y such that is a given value.
What shape is the locus? Do you notice any other interesting features?

Now look at and in our original diagram.

What do you see?

Those two angles are congruent, since they are both right angles!
So points F and D lie on the same circle as points B and P, just as points Y and Y' were on the same circle as points X and Z in our animation.

Now look at angles γ and σ. Again, just like the angles in our animation, they are angles which subtend a given segment (namely, ) and whose vertices lie on the same circle as the endpoints of that segment.

So γ = σ
(4)

Similarly, using points C, E, P, and F (see diagram below), we have
α = ε
(5)

Let's review what we know so far:

(1)	(β + γ) + δ = 180°
(2)	δ + (β + α) = 180°
(3)	γ = α
(4)	γ = σ
(5)	α = ε

By (3), (4), and (5), we have
ε = α = γ = σ

So ε = σ

So and are vertical angles.

Thus, D, E, and F are collinear.

Q.E.D.

We have shown that anytime the pedal point P is on the circumcircle of ΔABC, then we have that the pedal triangle is a degenerate triangle, or a Simson line.

Check out this GSP file to see what happens as we trace the [infinitely extended] Simson line as P travels around the circumcircle. Pretty cool, huh?

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