# The Proof of the Simpson Line Theorem

In this post, we explore the relationship between a point on each line containing the sides of a triangle that are nearest to a point on its circumcircle. In particular, consider point $P$ on the circumcircle of triangle $ABC$ as shown below.  We want to find a point on each line containing segments $AB$, $AC$ and $BC$ that are nearest to $P$.  To determine the location these points, we construct three lines through $P$ that are perpendicular to the three sides of the triangle. The intersections of the lines and the sides, namely $L$, $M$, and $N$ as shown below are the points nearest to $P$.

Notice that $AC$ is extended since the point perpendicular to $P$ is not on the segment. What do you observe? You might want to use GeoGebra, or any Geometry software, to construct and investigate the figure.

From the figure, it seems that the three points are collinear. Verify this conjecture using GeoGebra or any Geometry software. Using the Geometry software, move the vertices of the triangle and see if the conjecture is always true. Use the line through two points tool and draw a line through points $L$ and $M$. What do you observe? Are they always collinear?

Some Elementary Observations

Before further discussion, we take note of the following observations:

1. Notice that we can show that points $L$, $M$, and $N$ are collinear if we can show that $m \angle LMP + m \angle PMN = 180^\circ$.
2. Looking at the figure above, we observe that $ABPC$ is a cyclic quadrilateral.
3. Since $\angle PMB = 90^\circ$, By Thales’ theorem, points $P$, $M$, and $B$ are on the circle. In addition, $\angle PNB = 90^\circ$ which also means that point $N$ is on the same circle as those of $P$, $M$ and $B$. In effect, $PMNB$ is also a cyclic quadrilateral.
4. It follows from 3 that $PCML$ is also a cyclic quadrilateral.

We now use the observations above to prove our conjecture. We know that the measure of the opposite angles of a quadrilateral add up to $180^\circ$. As we have mentioned above, proving that \$latex L, M, and N are collinear, it is sufficient to show that $m \angle PMN + m \angle PML = 180^\circ$.

Proof

In quadrilateral $ABPC$, $m \angle ACP + m \angle ABP = 180^\circ$. Similarly, in quadrilateral $PMNB$, $\angle PMN + \angle ABP = 180^\circ$. It follows that $\angle PMN \cong \angle ACP$. In addition, in quadrilateral $PCML$,

$m \angle PMN + m \angle PML = m \angle ACP + (180 - \angle ACP) = 180^\circ$.

This is what we want to show.

The line above containing points $L$, $M$ and $N$ is called the Simson line named after Robert Simson, a Scottish mathematician. The theorem we just proved above is the Simpson Line Theorem.