Simson line

Dependencies: Thales converse (right angle $\Rightarrow$ on the circle with the hypotenuse as diameter), Opposite angles of a cyclic quadrilateral are supplementary (concyclic four points $\Leftrightarrow$ inscribed angles on the same chord equal / opposite angles supplementary), Linear pair sums to 180° (linear pair $= 180^\circ$).

Statement

Let $\odot O$ be the circumscribed circle of $\triangle ABC$, and let $P$ be any point on $\odot O$ (other than $A$, $B$, $C$). Drop perpendiculars from $P$ to the three lines containing the sides of $\triangle ABC$, with feet

Then the three feet $X$, $Y$, $Z$ are collinear — and this line is called the Simson line of $\triangle ABC$ with respect to $P$.

Conversely: if the three feet of perpendiculars from a point $P$ in the plane to the three sides are collinear, then $P$ must lie on the circumcircle. Here we prove only the forward direction; the converse is similar (push the same inscribed-angle relations back).