# Two Common Chords in Three Concurrent Circles

Points $N, P, Q$ are collinear. For a fourth point $M,$ consider circumcircles, $(A)$ of $\Delta MNP$, $(B)$ of $\Delta MPQ$, $(C)$ of $\Delta NPQ.$ $D$ a point on $(B)$, $F=DM\cap (A),$ $G=FN\cap (C).$ Prove that, if one of the segments $DP$, $FP$, $GQ$ is a diameter in the respective circle, so are the other two.

(The applet below illustrates the problem.)

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Solution ### Solution

Points $N, P, Q$ are collinear. For a fourth point $M,$ consider circumcircles, $(A)$ of $\Delta MNP$, $(B)$ of $\Delta MPQ$, $(C)$ of $\Delta NPQ.$ $D$ a point on $(B)$, $F=DM\cap (A),$ $G=FN\cap (C).$ Prove that, if one of the segments $DP$, $FP$, $GQ$ is a diameter in the respective circle, so are the other two.

If $DP$ is a diameter of $(B)$ then, as we know, $FP$ is a diameter of $(A),$ and vice versa. If $FP$ is a diameter of $(A)$ then angle $FNP$ is right. $FN\perp NP$, which is the same as $GN\perp NQ$ so that $GQ$ is a diameter of $(C),$ and vice versa. 