Three Overlapping Circles (Haruki's Theorem)

Haruki's Theorem: Given three overlapping circles & their intersections labelled as shown below, then (BC/CD)*(DE/EF)*(FA/AB) = 1.

Three Overlapping Circles (Haruki's Theorem)

Challenge: Can you explain why (prove) Haruki's theorem is true?

Hint: Try using Ceva's theorem and the power lines of a triangle to prove the result.

For a proof, read my paper in the Oct 2000 issue of the KwaZulu-Natal Mathematics Journal at Overlapping Circles.

Note: This delightful theorem has been named after Hiroshi Haruki (died September 13, 1997) from the Dept. of Mathematics, University of Waterloo, Canada.

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Michael de Villiers, created 13 July 2008; modified/adapted to WebSketchpad 30 August 2021.