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

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 .

EqualCirclesConic.html

Menelaus.html

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