Simson Line Theorem

Theorem
Let $\triangle ABC$ be a triangle.

Let $P$ be a point on the circumcircle of $\triangle ABC$.

Then the feet of the perpendiculars drawn from $P$ to each of the sides of $\triangle ABC$ are collinear.


 * Simson-line-1.png

This line is called the Simson Line.

Proof
In the figure above, construct the lines $BP$ and $CP$.


 * Simson-line-2.png

By the converse of Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $EPDB$ is cyclic.

By the converse of Angles in Same Segment of Circle are Equal, $EPCF$ is cyclic.

Therefore:

This gives:
 * $\angle DEP + \angle PEF = 180 \degrees$

Hence $DEF$ is a straight line.