Pedals and Arcs

by

Patty Wagner

My goal today is to convince you that that the pedal triangle of a pedal triangle of a pedal triangle of a point is similar to the original triangle. That is, ΔLMK is similar to ΔABC.

We begin by constructing the pedal triangle of point P with ΔABC. Recall that the pedal triangle is formed by constructing the perpendicular to each side of the triangle passing through point P. The points of intersection of the perpendicular with the sides of the triangle form the vertices of the pedal triangle. ΔDEF is our first pedal triangle.

We can construct the circumcircle of ΔDAF and see that ∠PAF inscribes arc PF. Notice that ∠FDP inscribes the same arc. Since inscribed angles in the same circle subtending the same arc are equal, we know that ∠PAF ≅ ∠FDP.

We mark these angles and construct our next pedal triangle:

By again constructing the circumcircle, we can see that ∠GHP ≅ ∠GDP for the same reasons as we saw above.

 

 

Continuing the pattern with the pedal triangle, ΔKLM, we get the following:

Since we are only concerned with our inner pedal triangle and how it relates to ΔABC, we can rid ourselves of the extraneous information and note that ∠CAP ≅ ∠KLP. We can call that angle, θ. Now we shift our focus and see that we can make the same argument for the other angles:

Angle BAP ≅ to angle DFP. Continuing in the same manner for every angle in our pedal triangles, we end up with:

If we remove all the extraneous information again and look simply at ΔABC and ΔKLM, we can see that we have two sets of corresponding angles that are congruent:

• α + θ = ∠BAC ≅ ∠MLK = α + θ
• γ + β = ∠ACB ≅ ∠LKM = γ + β

The Angle-Angle Postulate says that when two angles of one triangle are equal to two corresponding angles of the other triangle, the two triangles must be similar. Therefore, ΔABC ∼ ΔLMK.

Q.E.D.

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