The Degenerate Pedal

By Paulo Tan

The investigation starts with the construction of triangle ABC. An arbitrary point P is then chosen and the pedal triangle DEF is constructed (fig1).

Fig1.

Next you’ll notice that when the point P is located on point A, B, or C the vertices of the pedal triangle becomes collinear or in other words a degenerate triangle (fig2).

Fig2.

Click HERE to view the different pedal triangles formed by changing the position of P.

You may have also noticed that the pedal triangle is degenerate at other locations besides the vertices of triangle ABC (fig3).

Fig3.

Notice that point P lies on the circumcircle of triangle ABC (fig4). Obviously, points A, B, and C are on the circumcircle. Thus, it appears that if point P is on the circumcircle, then the pedal triangle is degenerate.

Fig4.

Click HERE to see pedal triangle when P is on the

circumcircle of triangle ABC

Let’s try to prove our hypothesis.

We’ll assume that point P is on the circumcircle of triangle ABC.

We also know that:

a) line PF is perpendicular to line AF,

b) line PE is perpendicular to line BE,

c) line PD is perpendicular to line CD.

Since quadrilateral BAPC is inscribed in a circle, then it is cyclic. If the quadrilateral is cyclic then opposite angles are supplementary (fig5).

Fig5.

When we construct segment BP, we’ll notice that segment BP is the hypotenuse of both triangle BFP and triangle BEP. Thus segment BP is the diameter of a circle with center G (fig6).

Fig6.

Since quadrilateral BFPE is inscribed in a circle then it is cyclic.

When we construct segment PE, we’ll notice that segment PE is the hypotenuse of both triangle PEC and triangle PDC. Thus segment PE is the diameter of a circle with center H (fig7).

Fig7.

Since quadrilateral DPCE is inscribed in a circle then it is cyclic. Notice that angle CPE and angle CDE are both inscribed angles that share a common intercepted arc.

When we construct segment AP, we’ll notice that segment AP is the hypotenuse of both triangle PDA and triangle PFA. Thus segment AP is the diameter of a circle with center J (fig8).

Fig8.

Once again, since quadrilateral AFPD is inscribed in a circle then it is cyclic. Notice that angle FDA and angle FPA are both inscribed angles that share a common intercepted arc.

Fig9.

Hence we proved our initial hypothesis that if point P is on the circumcircle then the result is a degenerate pedal triangle.

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