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!

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