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 \((B)\), \(F=DM\cap (A),\) \(G=FN\cap (C).\)

In three concurrent circles Two Common chords join end-to-end. Their configuration is closely related to the corresponding diameters - problem

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

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.

In three concurrent circles Two Common chords join end-to-end. Their configuration is closely related to the corresponding diameters - solution

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.

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