Let be any triangle. Now let be a point, which is not one of the vertices , , . Drop perpendiculars from to the three sides of the triangle. Next, label the intersection of the lines from with the sides , , , as , ,, respectively. Then the triangle is called the pedal triangle.
We will examine the special case where the point lies on the circumcircle of . In this case the feet of the perpendiculars drawn from to the sides , , are collinear, which is called the Simson Line. We will prove this is true.
Now, since is perpendicular to and is perpendicular to , we have that the point must lie on the circumcircle of .
We can employ similar arguments to show that lies on the circumcircle of as well as the circumcircle of .
From this we have that , , and are all cyclic quadrilaterals.
So from the fact that is a cyclic quadrilateral we have that,
which in turn tells us that
But is also a cyclic quadrilateral, and so we have
Now, opposite angles in a cyclic quadrilateral are supplementary and so it follows that
Next we need only subtract and we have
Recall that both and are cyclic quadrilaterals, and so
Now we can combine this with our previous result and we have that
Therefore, , , are collinerar. Q.E.D.
Furthermore, the converse is also true. That is, if the feet of the perpendiculars dropped from a point to the sides of the triangle are collinear, then is on the circumcircle.
Proof: Left to reader as an exercise!