Napoleon Converse

If GHI is an equilateral triangle, and three circles with centers at G, H and I are drawn so that they are concurrent in a point, then equilateral triangles on the sides of triangle ABC formed by the other intersections of circles I and G, G and H, and H and I, respectively, are inscribed in circles G, H and I.

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Napoleon General Converse

If GHI is any triangle, and three circles with centers at G, H and I are drawn so that they are concurrent in a point, and any triangles DBA, BEC and ACF are drawn on the sides of triangle ABC formed by the other intersections of circles I and G, G and H, and H and I, respectively, so that they are inscribed in circles G, H and I, then angle D = angle G, angle E = angle H, and angle F = angle I. (An obvious specialization is that if triangles DBA, BEC and ACF are constructed similar to GHI, then they are inscribed in circles G, H, and I).

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Can you generalize (and prove) the above result to a quadrilateral and four concurrent circles? Can you generalise further?

Specialization: Miquel's Theorem Converse

If GHI is any triangle, and three circles with centers at G, H and I are drawn so that they are concurrent in a point, and any point D on circle G is chosen, and line DB meets circle H again in E, line EC meets circle I again in F, then F, A and D are collinear, and DEF is similar to GHI.

Please enable Java for an interactive construction (with Cinderella).

Can you generalize (and prove) the above result to a quadrilateral and four concurrent circles? Can you generalise further?


For several more elaborate converses than those above, read John Wetzel's (1992) AMM paper "Converses of Napoleon's Theorem".


Created by Michael de Villiers, 22 Feb 2008 with Cinderella. Further edited, 10 Aug 2015.