Straightedge-only construction of a perpendicular

Solution: Taking it as known that the perpendicular bisectors of a triangle are concurrent, and that this implies the concurrence of the altitudes of a triangle, our construction and proof are simple.

Draw lines from the endpoints $E_1$ and $E_2$ of the circle through the point (call it $A$), meeting the circle in $B$ and $C$; now extend $E_1C$ and $E_2B$ until they meet in $D$. Now, by Thales' theorem, $E_1B$ and $E_2C$ are altitudes of $\triangle E_1DE_2$, meeting in the point $A$. Thus, the third altitude of this triangle, dropped from $D$, also passes through $A$; that is, $DA$ is the third altitude of the triangle. As such, it is perpendicular to the diameter of the circle.