Consider three chords AB, AC, AD of a circle from a point A. The three circles having diameters these chords intersect in three other points B*, C*, D* lying on the same line. This follows from Simson's (look at Simson.html ) theorem, since these points are feet of the perpendiculars from A to the sides of triangle BCD.
Consider three chords AB, AC, AD of a circle (c) from a point A. The three circles each with the property of its points to view the corresponding chord under the same (or complementary) fixed angle w have their second intersectin points B*, C*, D* on the sides of triangle ABC and also lying on the same line. Besides the centers of these circles lie on a circle passing through A and the center of (c). This is a generalization of the previous property.
From the definition of circles we have angle(AA*B)+angle(AA*C)=pi and analogous equations for the angles at B* and C*. This proves that {A*,B*,C*} are on the sides of triangle ABC.
angl(B*C*D)=angle(B*AD) by the cyclic quadrangle AC*DB* and angle(A*C*B)=angle(A*AB) by the cyclic ABA*C*. But triangles AB*D and ABA* are similar having by definition angle(DB*A)=angle(BA*A) and angle(ABA*)=angle(ADB*) by the cyclic ABCD. This proves the collinearity of {A*,B*,C*}.
The statement on centers is an easy exercise.
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1. Three diameters property ![[0_0]](ThreeDiameters_files/ThreeDiameters_0_0_0.png)
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