## Griffiths Points in Cyclic Quadrilateral

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Given a cyclic quadrilateral ABCD and a straight line l through its circumcenter, Griffiths' theorem supplies four Griffiths points for line l and four triangles ABC, BCD, CDA, DAB. As J. Tabov observed, the four points are collinear. In addition, the applet shows that the four pedal circles of any point on l with respect to the four triangles are concurrent. This is true for any complete quadrangle ABCD. For the cyclic quadrilateral, the point of concurrency lies on the line determined by the Griffiths points.

(The applet clearly shows that the 9-point circles of triangles ABC, BCD, CDA, DAB also concur. This result is treated elsewhere.)

Proof

### References

1. J. Tabov, Four Collinear Griffiths Points, Mathematics Magazine, v. 68, n 1, February 1995, pp. 61-64 ### Proof

I shall follow the convention used in establishing the existence of Griffiths points.

Assume there are four concyclic points A1, A2, A3, and A4, with the circumcenter at the origin and the circumradius of 1. Taken by three, they form four triangles Tm = ΔAiAjAk, where i, j, k, m are distinct elements from {1, 2, 3, 4}. To each Tm and a given line l there correspond Griffiths point Wm, m = 1, 2, 3, 4, which we found to be represented by complex numbers as

wm = (ai + aj + ak - ai aj ak) / 2.

To establish their collinearity we have to check that

(w1 - w2) / (w'1 - w'2) = (w1 - w3) / (w'1 - w'3) = (w1 - w4) / (w'1 - w'4).

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Define a = a1a2a3a4 for convenience. Easily

 w1 - w2 = (a2 + a3 + a4 - a2 a3 a4) / 2 - (a1 + a3 + a4 - a1 a3 a4) / 2 = (a1 - a2)(a3 a4 - 1) / 2.

Taking the conjugates gives

 w'1 - w'2 = (a'1 - a'2)(a'3 a'4 - 1) / 2 = (a1 - a2)(a3 a4 - 1) / 2a = (w1 - w2) / a.

which implies

(w1 - w2) / (w'1 - w'2) = a.

Obviously, the result will be the same if we replace w2 with w3 or w4. Thus indeed we have the condition of collinearity:

(w1 - w2) / (w'1 - w'2) = (w1 - w3) / (w'1 - w'3) = (w1 - w4) / (w'1 - w'4). 