Three Overlapping Circles (Haruki's Theorem)

Haruki's Theorem
Given three overlapping circles & their intersections labelled as shown below, then $$\frac{BC}{CD}\times \frac{DE}{EF}\times \frac{FA}{AB}=1.$$

Three Overlapping Circles (Haruki's Theorem)

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

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

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

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

Related Links
Experimentally Finding the Centroid of a Triangle with Different Weights at the Vertices (Ceva's theorem)
Balancing Weights in Geometry as a Method of Discovery & Explanation (incl. Ceva's theorem)
Power Lines of a Triangle
The Center of Gravity of a Triangle (Rethinking Proof activity - concurrency of medians, Ceva's theorem)

External Links
Ceva's theorem (Wikipedia)
Haruki's Theorem (Cut The Knot)
Haruki's Theorem (Wikipedia)
SA Mathematics Olympiad Questions and worked solutions for past South African Mathematics Olympiad papers can be found at this link.
(Note, however, that prospective users will need to register and log in to be able to view past papers and solutions.)

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