A Pedal Triangle and the Simson Line

 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.





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!