Generalization: Let such that are concyclic. Suppose are two circles through and respectively, such that they are corresponding circles between similar triangles . If , prove that are isogonals wrt .

Note: Another way to characterize is the property
Solution 1(Indirect):

Let denote the inversion transformation , then under .

Suppose . Let denote the isogonal conjugate of wrt and . We can show through angle chasing that . By more angle chasing we can obtain or . Since (by inversion's preservation of concylic points), therefore . Hence , so and are isogonals.

Solution 2(Direct, at last):

Lemma: Two circles intersect at two points . Given a point that satisfies and doesn't lie on . Prove that are isogonals wrt .

Proof: Let denote the foot of the bisector of on . Thus the -Apollonius circle of (aka the circle of similitude of passes through . Since , also bisects . Hence the lemma is proven.

Main proof: Since , therefore if , then lies on the circle of similitude of . By the Lemma above, this means are isogonals wrt . Becuase are isogonals wrt , therefore are isogonals wrt .