Problem
GEO-B3-M04-P008 Oblique Simson Line
B. New Original Problem. A point \(P\) lies on the circumcircle of \(ABC\). Through \(P\), lines are drawn meeting \(BC,CA,AB\) at the same directed angle \(\alpha\). Prove that the three intersection points are collinear.
C. Hint 1. Compare these points with the ordinary perpendicular feet.
D. Hint 2. They are obtained by one spiral homothety centered at \(P\).
E. Full Solution.
Let \(A_1,B_1,C_1\) be the ordinary perpendicular feet from \(P\). By the Simson theorem, they are collinear. The new points \(A_2,B_2,C_2\) on \(BC,CA,AB\) are obtained from \(A_1,B_1,C_1\) by the same change of angle at \(P\): from the perpendicular direction to the direction making angle \(\alpha\) with the side.
For each side, the correspondence \(A_1\mapsto A_2\), \(B_1\mapsto B_2\), \(C_1\mapsto C_2\) is given by one spiral homothety centered at \(P\) with angle \(90^\circ-\alpha\). Such a transformation sends a line to a line. Therefore \(A_2,B_2,C_2\) are collinear.
This is a useful generalisation, but directed angles must be handled carefully.