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.


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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

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