## Equal circles at vertices

Consider a triangle DEF and draw three equal circles with radius (r) centered at its vertices. The intersection points of the circles with the sides of DEF define three lines forming a triangle A'B'C' which is perspective to DEF.

To prove it, apply Menelaus theorem for the intersection points {D', E', F'} of the sides of DEF and corresponding sides of A'B'C'. By the condition examined in CeviansAndParallels.html this is equivalent in proving that:

Here [D,AB] denotes the distance of point D from line AB. Triangle DEF is the orthic of triangle ABC and A'B'C' is homothetic to it. In view of the relations of the angles: angle(F) = pi-2*angle(C), the typical factor of the above product becomes:

R is the circumradius of ABC. The other factors are obtained by cyclically permuting the letters A, B, C. Multiplying the three factors we get a fraction with numerator equal to the denominator, thus giving 1. Thus triangles DEF, A'B'C' are line-perspective and by Desargues theorem they are also point perspective.
This proposition implies that the six intersection points of triangles DEF and A'B'C' are on a conic. This is discussed in EqualCirclesConic.html .