**Assignment 9:**

** Exploring the Conditions in which the Vertices of a Pedal Triangle are Collinear**

by

Ángel M. Carreras Jusino

Goal:

- Show that the vertices of a pedal triangle are collinear when the pedal point lies in the circumcenter of the original triangle.

First lets define what is a

pedal triangle.Given a triangle

ABCand a pointP, the feet of the perpendiculars fromPto the sides ofABCform a triangle which is thepedal triangleforpedal point.PIn the following applet is shown a triangle

ABCa pedal point and its pedal triangle in yellow. Drag the pedal point to explore the pedal triangle for different locations of the pedal point in the plane. You can also drag the vertices of the triangleABCto modify it.

Exploration.Lets see what happen to the pedal triangle when the pedal point is located in the incircle, circumcircle, and nine point circle of triangle

ABC.Pedal point in the incircle.

Pedal point in the circumcircle

Pedal point in the nine point circle

Some observations about the pedal triangle when the pedal point is in one of the circles of triangle

ABC.

- When the pedal point is in the incircle the pedal triangle lie inside the triangle
ABC.- When the pedal point is in the circumcircle the vertices of the pedal triangle are collinear.
- When the pedal point is in the nine point circle one of the vertices of the pedal triangle always lies in the longest side of the triangle
ABC.From these observations, we will go into great detail in the 2

^{nd}.

Simson's LineIf the three vertices of the pedal triangle are collinear then the line that contains them is called the

Simson Line.Our conjecture is that if the pedal point lies in the circumcenter of the original circle then the vertices of the pedal triangle form the Simson Line.

Now we are going to demonstrate this conjecture.

Given a triangle

ABC,a pointP, and the pedal triangleDEFfor pedal pointP.Assume that

Plies in the circumcenter of triangleABC.Because

PD⊥BDandPE⊥BE,the pointPlies on the circumcenter of triangleEBD.Lets take some time to understand why this is true, because we are going to use this fact again.

Since

PDis perpendicular toBD, thenPBDis a right triangle with hypotenuseED.This hypotenuse can be used as a diameter of a circle containing point

P.

This is because the angle in a semicircle is always a right angle.Similarly because

PEis perpendiculars toBE, thenPEBis a right triangle with hypotenuseED.Since both triangles share the same hypotenuse, then both triangle can be inscribed in the circle with diameter

ED.So this circle contains the points

P,E,B, andD, therefore choosing any three points from these we can create a triangle which circumcircle contains the fourth point. In this case we choseE,B, andDfor the vertices of the triangle andPas the point lying in the circumcircle of such triangle.Because

PF⊥AFandPE⊥AE,the pointPlies on the circumcenter of triangleEAF.Because

PF⊥CFandPD⊥DC,the pointPlies on the circumcenter of triangleCDF.These implies that the quadrilaterals

PEBD,PEAF, andPCFDare cyclic quadrilaterals, i.e., quadrilaterals whose vertices lie in a circle. Note that also, by construction,ABPCis a cyclic quadrilateral.Now, since opposite angles of cyclic quadrilaterals are supplementary, we have:

(1) m∠

BED+ m∠PDB= 180°(2) m∠

PCF= 180°(3) m∠

AEP+ m∠PFA= 180°(4) m∠

ABP+ m∠PCA= 180°Since ∠

PCA≅ ∠PCF, substituting in (2) we get m∠PCA= 180°.Subtracting this result from (4), we have m∠

ABP.Now, given that m∠

ABPand m∠PBEare a linear pair, then m∠ABP =180° - m∠PBE.Therefore

m∠

PBE(5) m∠

PBE= 180°Now, since ∠

PBEand ∠PDEare inscribed in a circle, and they intercept a common arc, then ∠PBE≅ ∠PDE.Combining these last result and (5), we have m∠

PDE= 180°.Therefore, ∠

EDFis a straight angle. Consequently, the vertices of the pedal triangleEDFare collinear.