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Classical

Simson’s Theorem

Perpendicular feet and Carnot’s equal-angle extension

Two cyclic quadrilaterals align the perpendicular feet of a point on the circumcircle. The same angle argument gives Carnot’s extension to oblique projections.

Methods Orthogonal projections · Cyclic quadrilaterals · Inscribed angles

The theorem

Theorem

Let PP lie on the circumcircle of triangle ABCABC, and let D,E,FD,E,F be its perpendicular projections onto the lines BC,CA,ABBC,CA,AB, respectively. Then D,E,FD,E,F are collinear. Their line is the Simson line of PP with respect to ABCABC.

Triangle ABC with P on its circumcircle. The perpendicular feet D, E, F on BC, CA, and the extension of AB lie on one line.
Figure 1. The three perpendicular feet D, E, F lie on the Simson line.

Two cyclic quadrilaterals

Use directed angles modulo 180∘180^\circ, so that feet on side extensions are included.

The right angles show that P,D,C,EP,D,C,E and P,E,A,FP,E,A,F are cyclic: they lie on the circles with diameters PCPC and PAPA, respectively. Hence

∠DEP=∠DCP=∠BCP=∠BAP,\begin{aligned} \angle DEP&=\angle DCP\\ &=\angle BCP=\angle BAP, \end{aligned}

where the last equality uses the circumcircle of ABCABC, and

∠PEF=∠PAF=∠PAB.\angle PEF=\angle PAF=\angle PAB.

Therefore

∠DEF=∠DEP+∠PEF=∠BAP+∠PAB=0.\begin{aligned} \angle DEF &=\angle DEP+\angle PEF\\ &=\angle BAP+\angle PAB=0. \end{aligned}

Thus D,E,FD,E,F are collinear.

Carnot’s extension

The right angle can be replaced by any fixed angle θ\theta, with 0∘<θ<180∘0^\circ<\theta<180^\circ, measured with the same orientation on all three sides. Take D′∈BCD'\in BC, E′∈CAE'\in CA, and F′∈ABF'\in AB such that the directed angles from PD′,PE′,PF′PD',PE',PF' to the corresponding sidelines are all θ\theta.

The equal-angle condition still makes P,D′,C,E′P,D',C,E' and P,E′,A,F′P,E',A,F' cyclic. The same calculation therefore proves that D′,E′,F′D',E',F' are collinear. Taking θ=90∘\theta=90^\circ recovers Simson’s theorem.