## No-Pedal Collinearity

The title of the page may raise brows. The starting point for a short investigation that led to the statement illustrated by the applet below was Griffiths' theorem that states that pedal circles of the points on a fixed line through the circumcenter of a triangle are all concurrent in a single point - Griffiths' point corresponding to the given line and the triangle.

Hubert Shutrick found that the three circles corresponding to the intersections of the line with the sides of the triangle share an additional point and are thus coaxal: their centers are collinear. The applet illustrates a further generalization Dr. Shutrick has come up with: the line need not pass through the circumcenter and the collinear points need not be the centers of the pedal circles.

More accurately: let a transversal meet the side of ΔABC in points P (on BC), Q (on AC), R (on AB). Let O be a point not on a side of the triangle (and not necessarily on the transversal PQR) and AA', BB', CC' be the cevians through O. Denote X the intersection of AP with B'C', Y the intersection of BQ and A'C', Z the intersection of CR and A'B'. Then points X, Y, Z are collinear.

The final generalization does not mention pedal circles that were a main feature in Griffiths' theorem. This explains the caption.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

(In the applet the transversal can be translated or rotated if dragged next to the border of the applet area.)

Proof Copyright © 1996-2018 Alexander Bogomolny

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Choose homogeneous coordinates so that A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1) and O = (1, 1, 1), and let the line PQR be x/a + y/b + z/c = 0.

In this system of coordinates the side line AB is given by z = 0, whilst the cevian AA' corresponds to y = z.

By checking that the coordinates satisfy the equations of the lines given in the applet you find: the points A', B', C' are (0, 1, 1), (1, 0, 1) and (1, 1, 0), respectively; P = (0, b, -c), Q = (a, 0, -c) and R = (a, -b, 0); X = (b - c, b, -c), Y = (a, a - c, -c) and Z = (a, -b, a - b). A calculation shows that the determinant with rows X, Y and Z is zero so the points are collinear. The calculation is easy if you first take the sum of the first two columns from the third and then add half the third column to the other two. Copyright © 1996-2018 Alexander Bogomolny

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