Problem — Parallel chords from the endpoints of a diameter

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Let $MN$ be a diameter of a circle with centre $O$. Through $M$ draw a chord $MP$ and through $N$ a chord $NQ$ such that $MP \parallel NQ$. Prove that the two chords are congruent.

(original figure: the diameter $MN$ with the two parallel chords leaving from its opposite endpoints)

Solution

Let $H$ and $K$ be the feet of the perpendiculars drawn from the centre $O$ to the chords $MP$ and $NQ$ respectively. Since $MP \parallel NQ$ and the line $MN$ is a transversal, the angles $\widehat{PMN}$ and $\widehat{QNM}$ are alternate interior angles, hence congruent. Consider the right triangles $OHM$ and $OKN$: they have $\overline{OM} = \overline{ON} = r$ (radii), congruent acute angles $\widehat{HMO} = \widehat{KNO}$, and right angles at $H$ and $K$. By the congruence criterion for right triangles they are congruent, so $\overline{OH} = \overline{OK}$. Chords equidistant from the centre are congruent; therefore $\overline{MP} = \overline{NQ}$. $\blacksquare$