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 .
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nonuser
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Rishabh Shetty
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1Please, don't change the question. That trivializes my answer. If you want a new question then ask it again in a new thread. – nonuser Mar 23 '19 at 08:18
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Take a look at the picture. All those chords are paralell and the midpoints are on a common perpendicular bisector. 
nonuser
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@ Maria Mazur how can we prove that the above statement is true – Rishabh Shetty Mar 23 '19 at 08:08
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1You didn't ask that, you ask to help you understand the question, not to prove it for you. That is another question. – nonuser Mar 23 '19 at 08:12
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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.
Oscar Lanzi
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