Show that the locus of midpoints of a family of parallel chords of a circle is a diameter perpendicular those chords.

Question

Show that the locus of the midpoint of a family of parallel chords of a circle is a diameter which is perpendicular to the given family of chords.

Please help me understand the question with a figure .

Answer 1

Take a look at the picture. All those chords are paralell and the midpoints are on a common perpendicular bisector.

Answer 2

Let one parallel chord be $AB$ with midpoint $M$ and another one be $CD$ with midpoint $N$. Draw the perpendicular bisector of $AB$, then also draw the transversal $BC$ and its perpendicular bisector.

These two perpendicular bisectors intersect at the center $O$ of the circle, so $OC$ is congruent with $OD$ meaning the perpendicular bisector of $AB$ (which passes through $M$) is also the perpendicular director of $CD$ (which passes through $N$). Thereby the midpoints of both parallel chords must lie on the same line passing through the center $O$, QED.