In a circle, two chords CC' and DD' are drawn perpendicular to a diameter AB. Prove that the segment MM' joining the midpoints of the chords CD and C'D' is perpendicular to AB.

Let $AB$ be a diameter of the circle with centre $O$. Since $CC' \perp AB$, the diameter $AB$ is perpendicular to chord $CC'$; hence by Theorem

$AB$ passes through the midpoint of $CC'$. Similarly, $AB$ passes through the midpoint of $DD'$.

Now, reflecting across the line $AB$ (a line of symmetry of the circle), $C \leftrightarrow C'$ and $D \leftrightarrow D'$.
So, the chord $CD$ maps to the chord $CD'$ under reflection in $AB$.

The midpoint $M$ of $CD$ maps to the midpoint $M'$ of $CD'$ under this reflection. A point and its reflection across a line are joined by a segment perpendicular to that line (the line is the perpendicular bisector of the segment joining them).

Therefore, $MM' \perp AB$.