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symmedian.md

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Construction of the Symmedian

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.

Construction of the Isogonal Conjugate

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 .

Other proofs can be found here (section 3) and here.