Adams' Circle

Assume the incircle of DABC touches the sides BC, AC and AB in points D, E, and F respectively. The lines AD, BE and CF meet at the Gergonne point G of the triangle. DEF is known as the Gergonne triangle (and also contact triangle) of DABC. Suppose three lines are drawn through G parallel to the sides of the Gergonne triangle. These meet the sides of DABC in 6 points P, Q, R, S, T, U, as shown in the applet below. The following statement is due to C. Adams (1843):

  Six points P, Q, R, S, T, U are always concyclic. Moreover, the circle they lie on is centered at the incenter.

What we want to show is that the six points P, Q, R, S, T, U are equidistant from I. Since the points D, E, F are the pedal points (the feet of the perpendiculars from) of the incenter I, this is equivalent to having all the segments DP, DQ, ER, ES, FT, FU equal.



What if applet does not run?

AE and AF being tangents from A to the incircle, AE = AF. Hence, DEAF is isosceles. By construction, UR||EF, which implies that the triangles UAR and EAF are similar. So that, the former is also isosceles. Taking the difference of their equal side, we obtain

(1) ER = FU.

Similarly, DQ = FT and DP = ES.

Assume a line through A parallel to BC meets DE in X and DF in Y. DEAX is similar to DDCE, which is isosceles. Thus DEAX is also isosceles and

  AX = AE.


  AY = AF.

However, AE = AF, which implies that A is the midpoint of XY. Extend PS and QT to meet XY in M and N, respectively. The side lines of the triangles DXY and GMN are parallel, which makes the triangles similar and A the midpoint of MN. Subtracting gives

(2) MX = AX - AM = AY - AN = NY.

But MX = DP as the opposite sides of parallelogram DPMX. Similarly, NY = DQ. Together with (2), this gives

(3) DP = DQ.

Taking into account (1) completes the proof.

A beautiful result in triangle geometry.


  1. R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995, pp. 69-72


Copyright © 1996-2018 Alexander Bogomolny


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