Let be a triangle. Construct a point by intersecting the
tangents from and to the
circumcircle. Then is the symmedian of through vertex A.
First proof. Let the reflection of across the angle bisector of meet at . Then: \frac = \frac\frac=1
Second proof. Define as the
isogonal conjugate of . It is easy to see that the reflection of about the bisector is the line through parallel to . The same is true for , and so, is a parallelogram. is clearly the median, because a parallelogram's diagonals bisect each other, and is its reflection about the bisector.
Third proof. Let be the circle with center passing through and , and let be the
circumcenter of . Say lines intersect at , respectively. Since , triangles and are similar. Since :\angle PBQ = \angle BQC + \angle BAC = \frac{\angle BDC + \angle BOC}{2} = 90^\circ, we see that is a diameter of and hence passes through . Let be the midpoint of . Since is the midpoint of , the similarity implies that , from which the result follows.
Fourth proof. Let be the midpoint of the arc . , so is the angle bisector of . Let be the midpoint of , and It follows that is the
inverse of with respect to the circumcircle. From that, we know that the circumcircle is an
Apollonian circle with
foci . So is the bisector of angle , and we have achieved our wanted result. ==Tetrahedra==