扩展：等角共轭

Before reading this post, please read this [post][https://www.expii.com/e/640/]. (Just for fun: I had just proved the Trig Ceva in a different way.)

Given a triangle $ABC$ and $3$ cevians each from different vertex, given that the cevians are concurrent at $P$. Reflect each cevian from their respective angle bisectors. Then the $3$ new lines are also concurrent at a point $Q$, which we call it the isogonal conjugate of $P$.

Solution: Note that $\angle PAC=\angle P'AB, \angle PAB=\angle P'AC$, $\angle PBA =\angle P'BC, \angle PBC=\angle P'BA$, $\angle PCB=\angle P'CA$ and $\angle PCA=\angle P'CB$. So, by Trig Ceva theorem, we have $$\frac{\sin \angle P'AB}{\sin \angle P'AC}\cdot\frac{\sin \angle P'BC}{\sin \angle P'BA}\cdot \frac{\sin \angle P'CA}{\sin \angle P'CB}=\frac{\sin \angle PAC}{\sin \angle PAB}\cdot \frac{\sin \angle PBA}{\sin \angle PBC}\cdot \frac{\sin \angle PCB}{\sin \angle PCA}=1$$ and we are done by the converse of the Trig Ceva theorem. Yay.

Interesting Facts: The isogonal conjugate of the incentre $I$ is itself. The isogonal conjugate of the orthocentre $H$ is the circumcentre $O$. The isogonal conjugate of the centroid $G$ is (by definition) the symmedian point $K$. The isogonal conjugates of the Fermat points are the isodynamic points. The Brocard points are isogonal conjugates of each other. Try to prove each of them!

Challenge: Let a triangle $ABC$ has side lengths $BC=a, CA=b, AB=c$. Prove that the barycentric coordinates of the isogonal conjugate of the point with barycentric coordinates $x : y : z$ is $(\frac{a^2}{x} : \frac{b^2}{y} : \frac{c^2}{z})$