Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Prove it Ireland

Segments $AB$ and $CD$ are parallel chords of a circle centre $O$ , and $P$ is any point in the plane other than $O$ . If $|PA| = |PB|$ , prove $|PC| = |PD|$ .

Solution

A point $X$ is on the perpendicular bisector of $AB$ iff $|XA| = |XB|$ . Therefore, the centre $O$ of the circle is on the perpendicular bisectors of $AB$ and $CD$ . As $AB \parallel CD$ , this means that the two chords have the same perpendicular bisector. If $|PA| = |PB|$ , $P$ must be on this common perpendicular bisector, hence we also have $|PC| = |PD|$ .

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