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.

This line is called the Simson Line.

Proof

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

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.

Work In Progress In particular: Looking for the pages for the above converses and the (simple) unlinked theorem below. The theorem can be split trivially into Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles and Two Angles on Straight Line make Two Right Angles, but the same cannot be said for those converses. Category:Cyclic Quadrilaterals is of no help You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by completing it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{WIP}} from the code.

Therefore:

\(\ds \angle DEP\)

\(=\)

\(\ds \angle DBP\)

Angles in Same Segment of Circle are Equal: $EPDB$ is cyclic

\(\ds \)

\(=\)

\(\ds \angle ACP\)

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle: $ABPC$ is cyclic

\(\ds \)

\(=\)

\(\ds 180 \degrees - \angle PEF\)

Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles: $EPCF$ is cyclic

This gives:

$\angle DEP + \angle PEF = 180 \degrees$

Hence $DEF$ is a straight line.

$\blacksquare$

Source of Name

This entry was named for Robert Simson.