# Gayle Gilbert & Greg Schmidt

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!