Viviani by Inversion
Ptolemy's theorem can be proved by inversion from a simple identity AB + BC = AC, where point B is between A and C and all three are collinear. By exactly same reasoning we can prove another nontrivial statement:
Let point M lie on the arc between vertices A_{1}A_{n} of the circumcircle of a regular ngon
(1) 

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(1) 

Solution
Perform inversion f with center M and radius 1. Let
B_{j}B_{j+1} = A_{j}A_{j+1}/MA_{j}MA_{j+1} = a/d_{j}d_{j+1},
where a is the side length of the given ngon.
Points B are all collinear stretching successively from B_{1} to B_{n}, implying that
B_{1}B_{2} + B_{2}B_{3} + ... + B_{n1}B_{n} = B_{1}B_{n}.
Substitution now gives the desired identity (1).
Note: For n = 3, we may multiply by the product d_{1}d_{2}d_{3} to obtain
d_{3} + d_{1} = d_{2},
which is Viviani's Theorem.
References
 I. M. Yaglom, Geometric Transformations IV, MAA, 2009
Contact Front page Contents Geometry Up
Copyright © 19962017 Alexander Bogomolny