We can start with a triangle and its circunscribed circle. Given a point P on the circumcircle of a triangle, the feet of the perpendiculars from P to the three sides all lie on a straight line (Simson line or Simson-Wallace line)
We are going to see this property using this notation:
We have taken P to lie on the arc AC that does not contain B. Other cases can be derived by re-naming A, B, C.
If we can prove that these two angles are equal then points A', B', C' will be collinear.
We can use a consequence of a circle property (Euclides, III.21 or III.22) that saids that the opposite angles of every convex cuadrangle inscribed in a circle are together equal to two right angles.
Two right triangles are similar, then:
Points A, B', P, C' lies on a circle:
And points B',A',C,P lies on a circle:
Then points A', B', C' are collinear. This is called Simson Line or Simson-Wallace Line of P.
Coxeter, H. S. M. Introduction to Geometry, 2nd ed. New York: John Wiley and sons, 1969.
Coxeter, H. S. M. and Greitzer, S. L. Geometry Revisited. Washington, DC: Math. Assoc. Amer.
de Guzmán, Miguel 'The envelope of the Wallace-Simson lines of a triangle. A simple proof of the Steiner theorem on the deltoid'. RACSAM, vol. 95, 2001.