On the circle with diameter $AB$, the point $M$ was selected and fixed. Then the point ${{Q} _ {i}}$ is selected, for which the chord $M {{Q} _ {i}}$ intersects $AB$ at the point ${{K} _ {i}}$ and thus $ \angle M {{K} _ {i}} B <90 {} ^ \circ$. A chord that is perpendicular to $AB$ and passes through the point ${{K} _ {i}}$ intersects the line $B {{Q} _ {i}}$ at the point ${{P } _ {i}}$. Prove that the points ${{P} _ {i}}$ in all possible choices of the point ${{Q} _ {i}}$ lie on the same line. (Igor Nagel)