Construct a point D by intersecting the tangents from B and C to the circumcircle of triangle ABC, then AD is the symmedian of the triangle ABC, which implies ∠BAM = ∠CAD, where AM is the median.
Let's put the circumcenter onto the origin, rotate the triangle to make BC parallel to y-axis, then we get the equations of the circumcircle and line AD:
We can solve intersection E by eliminating y:
Here we use Vieta's formula because we already know a root xA:
Reflect E about x-axis at E' to make ∠BAE' = ∠CAE, then we have and
After some calculations, we get:
which implies AME' are collinear, so ∠BAM = ∠CAD. □
Other proofs can be found here.
P is an arbitrary point in triangle ABC. Construct the pedal trangle DEF and its circumcenter M. Reflect P about M at point Q, then Q is the Isogonal Conjugate of P.
Here proves but not
, because P may be a point outside the triangle, where we may have
instead of
.