Problem two states:
(2a) Let ABC be any triangle. Then if P is any point in the plane, the pedal triangle is formed by constructing perpendiculars to the sides of ABC (these are three points A', B', C' that are the intersections).
(2b) What if the pedal point P is the incenter of triangle ABC?
(2c) What if the pedal point P is the circumcenter of triangle ABC?
(2d) What if the pedal point P is the orthocenter of triangle ABC?
(2e) Interesting exploration, △ABC ≈ △A'''B'''C'''.
(2a) Let ABC be any triangle. Then if P is any point in the plane, the pedal triangle is formed by constructing perpendiculars to the sides of ABC (these are three points A', B', C' that are the intersections).
In order to establish a pattern, I first look at a few examples. After constructing triangle ABC, I used the Pedal Point GSP script (in assignment 5) to choose Pedal Point P.
Example 1: Pedal Point P is outside triangle ABC
In the above figure, the original triangle is shaded blue. After choosing a random Pedal Point P outside triangle ABC, we construct perpendiculars through P to the sides of triangle ABC.These perpendicular lines are colored orange in the above figure. The intersection of the line perpendicular to side BC, AC, and AB is A', B', and C', respectively. Now, A'B'C' is our pedal triangle (shaded green in the above figure). When the pedal point P is outside triangle ABC, A'B'C' is outside ABC as well.
Example 2: Pedal Point P is inside triangle ABC
We see that when the Pedal Point P is inside triangle ABC, the Pedal triangle A'B'C' is also inside triangle ABC.
(2b) What if the pedal point P is the incenter of triangle ABC?
I first start with constructing triangle ABC and then constructing the incenter (can use the Incenter script from assignment#5).
Recall that finding the incenter of triangle ABC is, by definition, finding the point of concurrency of the angle bisectors of each vertex. Since we are exploring the situation where the Pedal Point P is the incenter, we label this point of concurrency P. Notice that each angle bisector also bisects the line opposite its angle. As a result, finding the angle bisector of the angle at vertex A, the line constructed also bisects BC. The same can be said of the angle bisectors of vertex B to AC and vertex C to AB.
Next we construct perpendiculars to the sides of triangle ABC through P and label the foots of intersection of BC, AC, and AB as A', B', and C', respectively.
The lines of the angle bisectors are concurrent at the incenter, point P. The perpendiculars through point P to the sides of triangle ABC is represented in the above figure.
Thus, the pedal triangle A'B'C' is represented in the figure below.
Since P is the incenter of triangle ABC, we can construct the incircle of triangle ABC.
Notice the vertices of the pedal triangle, triangle A'B'C', all lie on the incenter of triangle ABC. As a result, the incircle of triangle ABC is the circumcircle of triangle A'B'C'. This implies the Pedal Point P is also the circumcenter of triangle A'B'C'.
(2c) What if the pedal point P is the circumcenter of triangle ABC?
To consider P as the circumcenter of triangle ABC, first construct triangle ABC and find its circumcenter (may use this script tool in assignment#5).
Recall that to find the circumcenter of triangle ABC, bisect the sides of triangle ABC to find the midpoint of each side.Then construct the perpendiculars to each midpoint. Specifically, through the midpoint of side AB, construct a perpendicular line, i.e. its perpendicular bisector. These three perpendicular bisectors of triangle ABC are concurrent at the circumcenter of triangle ABC. Label this circumcenter of triangle ABC as point P.
Now construct the perpendiculars to the sides of triangle ABC through P and label the foots of these perpendiculars.
Notice that the perpendicular bisectors of triangle ABC from the first step is the same as the perpendiculars constructed in the second step.
The pedal triangle is represented in the figure below.
We know the midpoints of triangle ABC are the vertices of the pedal triangle, triangle A'B'C.Thus, triangle A'B'C' is the medial triangle of triangle ABC.
Looking at triangle A'B'C', observe the altitudes of this triangle.
The altitudes of triangle A'B'C' are concurrent at the pedal point P. So, P is the orthocenter of triangle A'B'C'. Therefore, when Pedal Point P is the circumcenter of triangle ABC, P is also the orthocenter of the pedal triangle.
(2d) What if the pedal point P is the orthocenter of triangle ABC?
First, construct triangle ABC and then construct the orthocenter of triangle ABC (can use the orthocenter script from assignment#5).
Recall that to find the orthocenter of triangle ABC, find the altitudes of triangle ABC. Then the point of concurrency of these altitudes is the orthocenter. To find the altitudes, construct perpendicular lines through a vertex to it's opposite side. For example, construct a perpendicular line through vertex A to its opposite side BC. These lines intersec at the foot of the perpendicular. Thus the altitude is defined as the distance from the vertex A to the foot of the perpendicular. The same construction can be done for vertices B and C. A pictorial representation of the altitudes and foots of perpendiculars is below.
These three altitudes of triangle ABC are concurrent at its orthocenter. Label the orthocenter as the Pedal Point P.
Now, construct the perpendiculars through the Pedal Point P to sides BC, AC, and AB and label the points of intersections as A', B', and C', respectively.
Notice that the altitudes of triangle ABC from the first step is the same as the perpendiculars constructed in the second step.
The pedal triangle is shown below.
Since the pedal triangle was constructed using the altitudes of triangle ABC, the resulting triangle A'B'C' is the orthic triangle. Note that it is the orthic triangle as long as triangle ABC is acute.
Observe one last interesting detail of this pedal triangle by first constructing the angle bisectors of triangle A'B'C' and finding its incenter.
Notice that when the Pedal Point P is the orthocenter of triangle ABC, it is also the incenter of the pedal triangle.
(2e) Interesting extension, △ABC ≈ △A'''B'''C'''.
Start with the pedal point P as the incenter of triangle ABC. Then construct the pedal triangle, triangle A'B'C'.
From our previous explorations, we see that since the Pedal Point P of triangle ABC is the incenter, P is also the circumcenter of triangle A'B'C'. Now with P as the circumcenter of triangle A'B'C', create the pedal triangle to triangle A'B'C'.
From our previous explorations, we see that since the Pedal Point P of triangle A'B'C' is the circumcenter, P is also the orthocenter of triangle A''B''C''. Now with P as the orthocenter of triangle A''B''C'', create the pedal triangle to triangle A''B''C''.
From our previous explorations, we see that since the Pedal Point P of triangle A''B''C'' is the orthocenter, P is also the incenter of triangle A'''B'''C'''. We are now back to P as the incenter of a triangle, so conclude that △ABC ≈ △A'''B'''C'''.
