Assignment 9:

Pedal Triangles

by

Jenny Johnson

**Constructing
a Pedal Triangle**

First,
we are going to construct a triangle with vertices A, B, and C. Then we make an arbitrary point P and construct
a perpendicular line from P to each line containing sides AB, BC, and AC. Since the point P could be outside the
triangle, the perpendicular lines might also be found outside the triangle. That is why we construct the
perpendicular to the lines containing the sides instead of the sides
themselves. The three points at
which the perpendiculars intersect the lines containing the trianglesÕ sides
(labeled R, S, T) can be formed into a triangle. This triangle is called the Pedal Triangle. In the picture below, the red segments
are the sides of the triangle, the green lines are the lines containing the
sides, the purple dashed lines are the perpendiculars constructed from point P,
and the blue triangle is the pedal triangle.

If we move the point P outside triangle ABC, the pedal
triangle is altered, as shown in the pictures below.

For
your explorations with the pedal triangle, click here. This GSP file contains a pedal triangle
script tool.

Now,
we construct the pedal triangle of the pedal triangle and call it triangle
HIJ. Then we find the pedal
triangle of triangle HIJ, which is triangle KLM. These constructions can be seen in the picture below.

Without the construction lines, we get the following
picture.

**How is
triangle KLM related to the original triangle ABC?**

** **First, I measured each of angles of the original
triangle ABC and each of the angles of triangle KLM on GeometerÕs
Sketchpad. I then moved point P
and the vertices A, B, and C and found that angle A was always congruent to
angle KLM, angle B was always congruent to angle MKL and angle C was always
congruent to angle KML. This, of
course, is not a proof that the triangles are similar, but it helps me see that
I can find a way to prove the triangles are similar.

**Can we prove
that ****D****KLM is similar to ****D****ABC?**

** **LetÕs first look at the original triangle ABC with
point P and the pedal triangle RST.
We see that the quadrilateral PSBR has two right angles at vertex S and
vertex R.

This
creates two right triangles PSB and PRB. Based on what we know about circles, we can make a few
conjectures. There is a theorem
called the inscribed angles theorem that states that any angle θ inscribed
in a circle is half the measure of the central angle 2θ and the
intercepted arc 2θ. A
corollary of this theorem is that a right inscribed angle intercepts the circle
at the endpoints of the diameter. The angle measure does not change as its
vertex is moved to different positions on the circle. In our picture, this would mean that we can form a circle
with PB as the diameter and point S would lie on the circle. Similarly, R would have to lie on the
same circle with diameter PB. We thus
can draw the circle with diameter PB.

Using
the inscribed angle theorem again, we see that angle PSR and angle PBR both
intercept the same arc PR.

This
means angle PSR and angle PBR are both half the measure of arc PR and therefore
equal to each other. LetÕs say the
measure of these two angles is a.

Similarly,
if we look at angles PRS and PBS, then we see that they both intercept arc
PS. Thus, they are also equal
angles and weÕll label them with the letter b.

Looking
at quadrilaterals PTCR and PTAS, we see the same relationship with inscribed
angles. Thus, we can label all the
angles that are equal as follows.

Now
letÕs zoom in on the pedal triangle TSR and its pedal triangle HIJ.

Using
the same reasoning as above, angle PHR and angle PJR both intercept a circle at
the endpoints of its diameter PR.
Thus, we can form a circle with diameter PR and we can expect H and J to
both lie on that circle.

Now
we can see that angles HRP and HJP intercept the same arc HP, so the measures
of their angles are equal. Since
the measure of angle HRP is d, we know angle HJP must be d as well.

Angles
PRJ and PHJ are also equal because they intercept the same arc PJ and both are
equal to the measure of b.

Looking
at quadrilaterals PITH and PISJ, we see the same relationship with inscribed
angles. Thus, we can label all the
angles that are equal as follows.

Now
letÕs zoom in on the pedal triangle HIJ and its pedal triangle KLM.

Using the same reasoning as above again, angle PLJ and
angle PMJ both intercept a circle at the endpoints of its diameter PJ. Thus, we can form a circle with
diameter PJ and we can expect L and M to both lie on that circle. Also, angles LJP and LMP intercept arc
PL and therefore will both equal d, and angles PLM and PJM intercept arc PM and
will equal e.

Applying
the intercepted arc theorem to the entire triangles, we find that the angle
measures of both triangles are as follows.

**So are the
triangles KLM and ABC similar triangles?**

LetÕs
look at the angle measures of the two triangles.

We
see that angle B has a measure of a + b.
Angle MKL in the pedal triangle of the pedal triangle of the pedal
triangle is also a + b. Thus,
angle B = angle MKL. Angle C has a
measure of d + c. Angle KML also
has a measure of c + d. Thus,
angle C = angle KML. Angle A has a
measure of e + f. Angle KLM also
has a measure of e + f. Thus,
angle A = angle KLM. Since the
three angles in triangle KLM are congruent to the three angles in triangle ABC,
we know that they are similar triangles.