A Property of the Line IO:
A Proof From The Book

A proof by an anonymous visitor for a major part of a problem I already discussed on three occasions is bound -- in my view -- to be from the mysterious Book conceived by the late Paul Erdös. The book which is written by the supreme being contains only the most beautiful of the proofs. It is quite short. It makes an adept use of a few features of triangle geometry that may not be altogether common but are still quite elementary and appear to fit together with uncommon adroitness.

As usual, let I and O be the incenter and circumcenter, respectively, of triangle ABC. Suppose angle C is 30°, and that the side AB is laid off along each of the other two sides to give points D and E so that

EA = AB = BD.

Prove that the segment DE is both equal and perpendicular to IO.

As was noted elsewhere, the fact that DE is perpendicular to IO is independent of the angle C; the fact that was overlooked in [Honsberger, pp. 199-201] where the orthogonality of the two lines was derived based on their congruence. The other two solutions lacked in symmetry (that would be appreciated by A. Einstein) with respect to the vertices of the given triangle even when handling the matter of perpendicularity. However, a previous investigation showed that such symmetry is not alien to the geometric configuration of the problem.

So we prove the perpendicularity below.

30 December 2015, Created with GeoGebra

If K, L, M on BC, CA, AB are feet of the external bisectors of the angles A, B, C, then

KB/KC·LC/LA·MA/MB = b/c·c/a·a/b = 1

and K, L, M are collinear by Menelaus' theorem.

CK = ab/(c - b),
CL = ab/(c - a) and
CK/CL = (c - a)/(c - b) = CD/CE,

so that triangles CDE, CKL are centrally similar and DE parallel to LM, same as KLM. Let Ia, Ib, Ic be the excenters of ΔABC opposite to A, B, C. There is an intimate relationship between ΔABC and its excentral ΔIaIbIc (not to mention that they are perspective from I implying, in particular, by Desargues' theorem the collinearity of K, L, M.) First of all, the angle bisectors IaA, IbB, IcC of ΔABC are altitudes of the excentral triangle, I is its orthocenter, O its 9-point circle center, and the reflection O' of I in O its circumcenter, O'OI its Euler line. Quadrilaterals IbIcBC, IcIaCA, IaIbAB are all cyclic because of the right angles. Thus

KB·KC = KIb·KIc,
LC·LA = LIc·LIa,
MA·MB = MIa·MIb,

due to the Intersecting Secants Theorem. Therefore KLM is radical axis of the circumcircles (O), (O') of the triangles ABC, IaIbIc, perpendicular to the center line O'O, same as O'OI. In conclusion, DE is parallel to KLM, which is perpendicular to O'OI.

Hubert Shutrick has observed that K is the radical center of the two circumcricles (O) and (O') and the circle BCIbIc for it's the intersection of the radical axes of the pairs (O)/ BCIbIc and (O')/BCIbIc. It therefore lies on the radical axis of (O) and (O'). Similarly, point L and M also belong to that axis. It follows that KLM is indeed the radical axis of (O) and (O') and, as such, is perpendicular to the line of centers O'OI.


  1. R. Honsberger, From Erdös To Kiev, MAA, 1996.

Radical Axis and Radical Center

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