Let ABC be the given triangle. Bisect AB at D and AC at E
(I.10). From D and E draw perpendiculars to AB and AC
respectively (I.11), meeting at F. Join FA, FB, FC. By I.4
applied to the two right triangles at D, FA=FB; similarly at
E, FA=FC. Thus FA=FB=FC, and the circle with centre F
and radius FA passes through all three vertices.