Theorem
The pedal circles of a point and its isogonal conjugate coincide.
Proof
The quadrilateral PBPACP is cyclic, for two opposite angle at BP and CP both measuring 90° add up to 180°. Therefore two angles PBPCP and PACP that subtend the same chord PCP are equal. Similarly, ∠QABQ = ∠QCQBQ. Since P and Q are isogonal conjugate, we also have ∠PACP = ∠QABQ. Therefore, ∠PBPCP = ∠QCQBQ. Subtracting these from the equal angles PBPA and QBCA we obtain ∠CPBPBQ = ∠BQCQCP. The latter angles subtend the same segment BQCP, which proves that the four points BP, BQ, CP, CQ are concyclic - they lie on the same circle. The center of the circle is necessarily the point of intersection of the perpendicular bisectors to BPBQ and CPCQ, which is the midpoint of the segment PQ.
The above argument is completely symmetric with respect to the sides of ΔABC. So it follows that all six points AP, AQ, BP, BQ, CP and CQ lie on a circle centered at the midpoint of PQ.
As we know, if P lies on the circumcircle of ΔABC, Q is a point at infinity. In that case, the radius of the above circle is infinite, and the circle appears to be a line. Such that, if P lies on the circumcircle of ΔABC, the feet of the three perpendiculars from P to the sides of the triangle lie on that straight line. The line is known as the Simson line of P with respect to ΔABC.
References
- R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995, pp. 67-68
- V. V. Prasolov, Essays On Numbers And Figures, AMS, 2000, pp. 65-69
