Investigation of the Simson Line

by Asli Ersoz

In this write-up, I will explore some properties of the Simson Line. But before going there, I need to define pedal triangle and then the Simson line.

Let triangle ABC be any triangle. Take any point P in the plane of ABC, then construct perpendiculars to the sides of ABC (extended if necessary). Locate three points R, S, and T that are the intersections. Triangle RST is the Pedal Triangle for Pedal Point P.

Here is an example:

You can click here to explore the different pedal triangles formed when the point P is taken in different places.

Have you observed any conditions for P when the three vertices of the triangle are colinear?

This happens when P is taken on the circumcircle of ABC. The triangle is called a degenarate triangle when the three vertices are colinear and the line segment is called the Simson Line.

Here is the Simson Line:

You can click here to observe what happens as P moves around the circumcircle of ABC.

An important question to be asked is: Why are these three points colinear?

Here is the proof:

To be able to prove this, I need to show that the measure of the angle BSR is equal to the measure of the angle TSC.

First let's construct some segments that will be useful. Construct BP and PC.

Look at quadrilateral BSPR. It is cyclic, meaning that its vertices lie on a circle, because angle BRP and angle BSP add up to 180 degrees (They are both right angles.).

This tells us that angle BSR is equal to angle BPR because they see the same chord.

Then, look at quadrilateral CPST. It is also cyclic because right angles PST and PTC see the same side (If you draw the circumcircle of PSC, for PTC to be right angle, T must be on the same circle.).

This tells us that angle TSC is equal to angle TPC.

We will make use of two more cyclic quadrilaterals: ABPC (obvious) and ARPT (See the two right angles?)

These tell us that

BAC + BPC = 180

BAC + RPT = 180

So,

BPC = RPT

BPT + TPC = BPR + BPT

TPC = RPB

We already know that TPC = TSC and BPR = BSR.

Hence,

TSC = BSR

This completes the proof!

Here is an observation about the Simson Line:

If we connect P to the orthocenter H of the triangle ABC, the Simson Line bisects PH. Moreover, the midpoint J of PH lies on the nine-point circle of ABC.