Published in **Mathematics & Informatics Quarterly**, 6(3), 169-171, Sept 1996.

A dual to Kosnita's theorem

Michael de Villiers, Mathematics Education,

University of Durban-Westville, Durban 4000, South Africa

In the September 1995 issue of the *Mathematics and Informatics Quarterly*, the following interesting result was mentioned without proof in the column on Forgotten Theorems:

**Kosnita's Theorem**. The lines joining the vertices *A*, *B*, and *C *of a given triangle *ABC* with the circumcenters of the triangles *BCO*, *CAO,* and *ABO* (*O* is the circumcenter of triangle *ABC*), respectively, are concurrent.

On the basis of an often observed (but not generally true) duality between circumcenters and incenters (eg. see De Villiers, 1996), I immediately conjectured that the following dual to Kosnita's theorem might be true, namely:

**Kosnita Dual**. The lines joining the vertices *A*, *B*, and *C *of a given triangle *ABC* with the incenters of the triangles *BCO*, *CAO,* and *ABO* (*O* is the incenter of triangle *ABC*), respectively, are concurrent.

Investigation on the dynamic geometry program *Sketchpad* then confirmed that the conjecture was indeed true. I then found that both results could easily be proved by the following useful result for concurrency, a proof of which is given in De Villiers (1996). Interestingly, the point of concurrency for the special case with equilateral triangles on the sides is called the Fermat-Torricelli point.

**Fermat-Torricelli Generalization**. If triangles *DBA*, *ECB* and *FAC* are constructed on the sides of any triangle *ABC* so that angle *DAB* = angle *CAF*, angle *DBA* = angle *CBE* and angle *ECB* = angle *ACF*, then *DC*, *EA* and *FB* are concurrent (see Figure 1).

Figure 1

It should further be pointed out that the above result is also true when the triangles are constructed inwardly. As shown in Figure 2 we clearly have for Kosnita's dual that angle *DAB* = (angle *A*)/4 = angle *CAF*, angle *DBA* = (angle *B*)/4 = angle *CBE* and angle *ECB* = (angle *C*)/4 = angle *ACF*, and from the above result it therefore follows that *DC*, *EA* and *FB* are concurrent.

Figure 2

Figure 3

Kosnita's theorem follows a little less directly from the Fermat-Torricelli generalization. In this case, the "base triangle" is triangle *DEF* with *A*, *B* and *C* the outer vertices (see Figure 3). Since *DBOA* is a kite, we have angle *BDO* = angle *ADO*. But *DBEO* and *DOFA* are also kites. Therefore, angle *BDE* = (angle *BDO*)/2 and angle *ADF* = (angle *ADO*)/2 from which follows that angle *BDE* = angle *ADF*. In a similar fashion can be shown that angle *BED* = angle *CEF* and angle *CFE* = angle *AFD*. From the F-T generalization, it therefore follows that *DC*, *EA* and *FB* are concurrent.

Reference

De Villiers, M. (1995). A generalization of the Fermat-Torricelli point. *The Mathematical Gazette*, 79(485), July, 374-378.

De Villiers, M. (1996). *Some Adventures in Euclidean Geometry*. Durban: University of Durban-Westville.