Call the arbitrary point $A$, and the endpoints of the diameter $B$ and $C$. Create triangle $ABC$ and mark the intersections of line $AB$ and $AC$ with the circle as $E$ and $F$ respectively. If $AB$ is tangent to the circle, then $B=E$. Same for the case if $AC$ is tangent. Since $BF$ and $CE$ are perpendicular to sides of the triangle by inscribed angle theorems, we know that their intersection must be the orthocenter of $\Delta ABC$. Call their intersection $H$ Draw line $AH$. Since this is also an altitude, we have that $AH$ is perpendicular to $BC$ as wanted.