Then

a) Lines AK, BL, CM are all equal.

b) All these lines pass through a point F.

c) All the angles formed at F by these lines are 60 degrees.

d) The centers of the equilateral triangles form again an equilateral triangle UVW (see Napoleon.html ).

e) The point F minimizes the sum of distances f(P) = PA+PB+PC.

For (a): compare triangles AKC, LBC and show them equal. The second results from the first by turningit by an

angle of π/3 about C. This implies also that the angle of AK to BL is π/3. From the equality ofangles AFL and

ACL follows that AFCL is cyclic, this implying that KFC is also π/3, thus F is distinguished on BL by its

property to view BC under the angle BFC which is 2π/3. This implies easily all propertiesup to (d).

For (e): Take a point P, join with A, B and C. Turn the triangle APC about A by an angle of 60 degrees (figure

below). You want to minimize BP+AP+CP = BP+PQ+QL. But this is always shorter than the segment BL.

AK, BL, CM are again equal but this time intersect at a point F on the extension of AK, for A>2π/3.

Also point F in this case is viewing BC under an angle of measure 2π/3, whereas the other sides are seen

under an angle of &p;i/3.

the more usually drawn. Again segments {AK, BL, CM} have equal lengths and pass through the same point,

which is viewing the sides under π/3 or 2π/3 angles. The point F' thus defined is the

point of the triangle.

proven above (see Trilinears.html ).

Last cancelation of factors due to the equilateral ACL. Hence the trilinears of points F, and, working analogously

for F', are given by the ratios:

Thus, it is seen that {F,F'} are

the triangle (see Isodynamic.html and for the calculation of trilinears of {J, J'} see ApolloniusCircles.html ).

Trilinears.html

Isogonal_Conjugation.html

Isodynamic.html

ApolloniusCircles.html

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