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Geometry Problem 903: First Fermat Point

The figure shows an acute triangle ABC with three equilateral triangle erected externally on the sides. Prove that (1) AA', BB', and CC' concur in a point F called the Fermat point. (2)The angles AFB, BFC, and AFC are equal to 120 degrees. (3) AF + BF + CF is the minimum possible.
The Fermat point of a triangle, is the solution to the problem of finding a point F such that the total distance from the three vertices of the triangle to the point is the minimum possible.

Fermat Point

 
 

    

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