Problem
GEO-B3-M04-P006 Converse Simson Theorem
B. New Original Problem. For a point \(P\), the projections \(A_1,B_1,C_1\) onto the lines \(BC,CA,AB\) of triangle \(ABC\) are collinear. Prove that \(P\) lies on the circumcircle of \(ABC\).
C. Hint 1. Repeat the proof of the Simson theorem backwards.
D. Hint 2. Collinearity replaces the equality of angles at \(B_1\).
E. Full Solution.
As in the direct theorem, the right angles give cyclic quadrilaterals \(P,A_1,C,B_1\) and \(P,B_1,A,C_1\). The collinearity of \(A_1,B_1,C_1\) gives equality of the corresponding directed angles.
Transferring these angles to the circles with diameters, we get \(\angle PCA=\angle PBA\). This is the cyclicity criterion for \(P,C,A,B\). Hence \(P\) lies on the circumcircle of triangle \(ABC\).
Use as a criterion: if pedal points are collinear, the point is on the circle.