Three Common Chords in Three Concurrent Circles

When two circles intersect, the common chord through one of the intersections is the longest when its endpoints are diametrically opposite the other point of intersection in the two circles. This property is extended to three concurrent circles.

Circles \((A)\), \((B)\), and \((C)\) concur at point \(D\). Chords \(KL,\) \(LM\), and \(KM\) pass through the other points of intersection \(G,H,F\)

Two concurrent circles. Common chords form a triangle. All such triangles similar and more ...

Prove that, if one of the segments \(DK\), \(DL\), \(DM\) is a diameter in the respective circle, so are the other two.

(The applet below illustrates the problem.)

14 January 2015, Created with GeoGebra

Created with GeoGebra

This is a return to a configuration that we met several times in the past. The statement will actually work for any number of concurrent circles.

|Contact| |Front page| |Content| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny


Search by google: