Thales circle theorem extended to Mahalanobis geometry

Frank Nielsen

October 20, 2017

This column is also available in pdf: filename MahalanobisThalesTheorem.pdf

1 Thales’ theorem in Euclidean geometry

In planar Euclidean geometry, Thales’ theorem states that any triangle pqr circumscribing a circle with one pair (p,q) of antipodal points is necessarily a right triangle. A pair (p,q) of antipodal points of a smooth convex object is such that the tangent lines at p and q are parallel to each other. See Figure 1 for an illustration, and [3] for a historical account (Thales of Miletus, 624–546 BC).

Theorem 1 (Thales’ circle theorem) Any triangle circumscribed by a circle with one side being a diameter is right-angle.


Figure 1: Thales’ circle theorem: Triangle pqr is right-angle at r where [pq] is a diameter. (p,q) is a pair of antipodal points.

2 Thales’ theorem in Mahanalobis geometry

Let DA(p,q) denote the Mahalanobis distance between two points p and q, for a positive definite matrix A 0:

         ∘ ---------------
DA(p,q) =  (p- q)⊤A(p - q) = ∥p- q∥A.
When A = I is the 2 × 2 identity matrix, the Mahalanobis distance amounts to the Euclidean distance DE(p,q) = p - q= ∘(p-- q)⊤(p-- q).

A Mahalanobis circle [2] CA(c,r) of center c and radius r is defined as follows:

CA (c,r) = {x : DA(c,x) = r}.
A Mahalanobis circle has an ellipsoid (Euclidean) shape.

Let us generalize Thales’ theorem as follows:

Theorem 2 (Thales’ Mahalanobis circle theorem) Any triangle circumscribed by a Mahalanobis circle with one pair of points being antipodal is right-angle.

Proof: Consider the Cholesky decomposition of A: A = LL = UU with L (U = L) a lower triangular matrix (an upper triangular matrix, respectively) with positive diagonal elements. The Mahalanobis distance amounts to calculate an ordinary Euclidean distance on affinely transformed points x= Lx = Ux:

            ∘ -----------------
DA (p,q)  =    (p- q)⊤LL ⊤(p- q),

         =  DE (L⊤p,L⊤q ).

Thus a Mahalanobis circles CA transforms affinely to a Euclidean circle CE = CI, and antipodal pairs of points on CA remain antipodal in CE.

Two vectors u and v are perpendicular in the Mahalanobis geometry if and only if uAv = 0. That is, if uAv = uLLv = (Lu)Lv = uv= 0.

A triangle pqr circumscribing the Mahalanobis circle CA with (p,q) an antipodal pair in Mahalanobis geometry transforms into a triangle pqrcircumscribing the Euclidean circle C= {Lx  :  x CA(c,r)} with (p,q) an antipodal pair. Therefore pqris a right-angle triangle in Euclidean geometry, and:

       ′  ′ ⊤  ′  ′
     (q- r ) (p - r ) =  0,                             (1)
(L⊤ (q - r))⊤L ⊤(p- r)  =  0,                             (2)
   (q- r)⊤LL ⊤(p- r)  =  0,                             (3)

       (q- r)A (p- r)  =  0.                             (4)
Therefore, pqr is a right-angle triangle in Mahalanobis geometry.


Figure 2: Thales’ circle theorem in Mahalanobis geometry: Triangle pqr is right-angle at r where [pq] is an antipodal pair of points. Notice that Mahalanobis geometry is generally not conformal so that a Mahalanobis right-angle does not visualize as a Euclidean right-angle.

Note that Mahalanobis geometry is not conformal when A = λI (for λ > 0), the scaled identity matrix. Therefore angles are not preserved in Mahalanobis geometry: That is, a Mahalanobis right-angle cannot be visualized as a Euclidean right-angle in general.

Squared Mahalanobis distances are the only symmetric Bregman divergences [1]. But Thales’ theorem do not extend to other (asymmetric) Bregman divergences.


[1]   Jean-Daniel Boissonnat, Frank Nielsen, and Richard Nock. Bregman Voronoi diagrams. Discrete & Computational Geometry, 44(2):281–307, 2010.

[2]   Frank Nielsen and Richard Nock. On the smallest enclosing information disk. Information Processing Letters, 105(3):93–97, 2008.

[3]   Christoph J Scriba and Peter Schreiber. Geometry in the Greek-Hellenistic era and late antiquity. In 5000 Years of Geometry, pages 27–116. Springer, 2015.