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.
Proof:
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
and
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!