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 ABC and a point P, the feet of the perpendiculars from P to the sides of ABC form a triangle which is the pedal triangle for pedal point P.

In the following applet is shown a triangle ABC a 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 triangle ABC to 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.

1. When the pedal point is in the incircle the pedal triangle lie inside the triangle ABC.
2. When the pedal point is in the circumcircle the vertices of the pedal triangle are collinear.
3. 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 2nd.

Simson's Line

If 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 point P, and the pedal triangle DEF for pedal point P.

Assume that P lies in the circumcenter of triangle ABC.

Because PDBD and PEBE, the point P lies on the circumcenter of triangle EBD.

Lets take some time to understand why this is true, because we are going to use this fact again.

Since PD is perpendicular to BD, then PBD is a right triangle with hypotenuse ED.

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 PE is perpendiculars to BE, then PEB is a right triangle with hypotenuse ED.

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, and D, therefore choosing any three points from these we can create a triangle which circumcircle contains the fourth point. In this case we chose E, B, and D for the vertices of the triangle and P as the point lying in the circumcircle of such triangle.

Because PFAF and PEAE, the point P lies on the circumcenter of triangle EAF.

Because PFCF and PDDC, the point P lies on the circumcenter of triangle CDF.

These implies that the quadrilaterals PEBD, PEAF, and PCFD are cyclic quadrilaterals, i.e., quadrilaterals whose vertices lie in a circle. Note that also, by construction, ABPC is a cyclic quadrilateral.

Now, since opposite angles of cyclic quadrilaterals are supplementary, we have:

(1) m∠BED + m∠PDB = 180°

(2) m∠PDF + m∠PCF = 180°

(3) m∠AEP + m∠PFA = 180°

(4) m∠ABP + m∠PCA = 180°

Since ∠PCA ≅ ∠PCF, substituting in (2) we get m∠PDF + m∠PCA = 180°.

Subtracting this result from (4), we have m∠PDF = m∠ABP.

Now, given that m∠ABP and m∠PBE are a linear pair, then m∠ABP = 180° - m∠PBE.

Therefore

m∠PDF = 180° - m∠PBE

(5) m∠PDF + m∠PBE = 180°

Now, since ∠PBE and ∠PDE are inscribed in a circle, and they intercept a common arc, then ∠PBE ≅ ∠PDE.

Combining these last result and (5), we have m∠PDF + m∠PDE = 180°.

Therefore, ∠EDF is a straight angle. Consequently, the vertices of the pedal triangle EDF are collinear.