For this assignment we will be exploring the Simpson Line of pedal triangles. To begin this exploration let's review the definition of the pedal triangle and see what possibilities we may have for finding the Simson Line. We know that the pedal triangle ΔRST is formed by constructing perpendiculars to the sides of ΔABC with respect to a point P anywhere in the plane. The points R, S, and T are the intersection points of the perpendiculars drawn from the point P to the sides of ΔABC. The sides of ΔABC may be extended to complete the intersection.
Below we have are some examples of our pedal triangle. Notice when the point P lies on a vertice of ΔABC two of the vertices of ΔRST also lie on the same point as P. It should also be observed that ΔRST degenerates when P lies on a vertice and when P lies on ΔABC we can clearly see that ΔRST does not degenerate. Both of these are major clues for our search of the Simpson Line.
We can also see from our graphs that the pedal triangle also exists when the point P is a certain distance outside of ΔABC. Is it safe to assume that our range of where the Simson Line may exist somewhere between the sides of ΔABC and and the point of existence beyond the exterior of ΔABC? So where does this range reside? We know the Simson Line exists at each vertice of ΔABC which gives us three definite locations. We also know that the range is within a fairly close proximity to the exterior of ΔABC. So where is this range?
The truth of the existing Simson Line lies on the circumcircle for the circumcircle is the range we have been searching for. Let's now look at verifying this truth. The picture above will help us calculate angles using Thales' Theorem and properties of cyclic quadrilaterals in order to prove that R, S, and T are collinear when P lies on the circumcircle. We will first observe that we have three cyclic quadrilaterals: Quadrilateral ABCP, Quadrilateral RASP, and Quadrilateral PCTS. It is obvious that Quadrilateral ABCP is cyclic for we are given that all points A, B, C, and P lie on the circumcircle. CP and AP are the diameters of their corresponding circumcircle and are also the diagonals of their corresponding quadrilateral. By Thales' Theorem we know that ∠ARP and ∠PSA are right angles. We can verify this by using the opposite angles summation property of cyclic quadrilaterals to verify this. The summation of opposite angles in a cyclic quadrilateral equal to 180 degrees. We can also see by Thales' Theorem that ∠PSC and ∠CTP equal 90 degrees as well. By putting all of this information together we can prove that R, S, and T are collinear.
First, we showed that vertices R and S have equal angles and are the end points of the diagonal for the cyclic Quadrilateral RASP. Hence, R and S lie on a line. Next we must show that T also lies on this line. Let's recall that Quadrilateral PCTS is cyclic so segment ST forms a side of the quadrilateral. In addition, vertex T of the pedal triangle forms a right angle with side BC of ΔABC, i.e. T is an intersection point on the line BC. We now have that T intersects the cyclic Quadrilateral ABPC and it is a vertex of cyclic Quadrilateral PCTS. Recall also that the point C is a vertex for cyclic Quadrilateral ABPC AND for cyclic Quadrilateral PCTS. Therefore, no matter how C changes it will always be an element of the circumcircle of ∆ABC, it will always be the end point of the diameter of the circle with center C'', and as the diameter CP tends to zero the points S and T tend to the point P where the vertex C also tends. Even as C changes T will not be removed from its intersection and vertex position. And we know from the previous discussion above that our pedal triangle ∆RST degenerates when P is located at C, i.e. two vertices of the pedal triangle will also be located there and in this example those vertices would be S and T. This reasoning remains the same when vertices A, B, and P are moved as well. All of the cyclic quadrilaterals and circumscribes are connected by common intersection points and vertices and every change in the position of these points will effect the position of all points while maintaining the consistent relationship between all cyclic quadrilaterals and circumscribes. Therefore, All of our possible cases have been covered, and we can conclude that R, S, and T are collinear and form the Simson Line when the point P lies on the circumcircle of ∆ABC.
Click Here to See a GSP5 animation of the Simpson Line.
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