Pedal Triangle


By: Carly Cantrell



The Pedal Triangle does not necessarily have petals like flower petals, but they still are pretty and pretty special.


You can easily create a pedal triangle:


     First, create any triangle ABC.


     Then create a point inside of the triangle and label it P, this will become your pedal point.


     Next, construct three perpendicular lines, each through a segment and P.


     Construct points at the intersections of the perpendicular lines and triangle ABC.


     Once you have created these three points, and then make segments between the three points, called the pedal triangle.






Explore for yourself: Pedal Triangle





Simpson Lines:


The sides of triangle ABC can be extended into lines, which allows P, the pedal point, to move outside of triangle ABC. In particular locations, the pedal triangle collapses into a line segment. When this occurs, this line segment is called a Simpson Line. An example, in light blue, is shown below:








Prove: If P is any point on the circumcircle of triangle ABC, then the feet of the perpendiculars from P to the sides of the triangle are collinear.


Proof: Let E, F, G, be feets of perpendiculars from point P on the circumcircle.


Observe, H is the circumcenter of triangle ABC. So, A, B, C, and P are all concylic because P is placed on the circumcircle of triangle ABC.










<PFC is a right angle, by construction. Using Thales Theorem, we know the circumcircle of triangle PFB has a diameter the length of segment PC. Similarly, <PGC is also a right angle so it follows that the diameter of the circumcircle for triangle PTB is the length of the segment PC. These circumcircles have the same diameter, which ensues they are the same circle. Thus, P, F, G, and G are concyclic.












<PEB and <PFB are both right angles, by construction. Following the same method, P, F, B, and E are concyclic.












We know that the opposite angles of concyclic quadrilaterals are supplementary. So, we can observe that ABP + ACP = 180 and PEG + ACP = 180.

Hence, ABP and PEG are both supplementary to ACP, which suggests ABP = PEG.

Looking at quadrilateral BEPF, we see PBE = PFE.

Therefore, we are able to replace PFE with PBE to show that ABP + PFE = 180.

Furthermore, PFG + PFE = ABF + 180 - ABP.

This means that PFG and PFE are a linear pair, hence E, F, and G are colinear.






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