[alogo] Adams Circle of a triangle

Consider the Gergonne triangle t* = A*B*C* of triangle t = ABC, whose vertices are the contact points of the incircle (c) of t with the sides of t. Lines AA*, BB*, CC* concur at the Gergonne point G. Adams theorem (1843) states that the parallels to the sides of t* from G, intersect the sides of t in 6 points lying on a circle (d) concentric to the incircle c.

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The following remarks lead to the proof ( look R. Honsberger's book: Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA vol. 37, p.63).
- A'A* = B"B* because of the tancency of lines CB and CA at A* and B*. Analogously C'C* = A''A* and B'B* = C''C*.
- Draw B**C** parallel to BC from A. Triangles B*AC** and C*AB** are isosceli and AB** = AC* = AB* = AC**, i.e. A*A is a medial line of triangle A*B**C**.
- Previous result follows also from the basic facts about symmedians, since A*A is a [symmedian] of triangle A*B*C* and C**B** is antiparallel to side B*C* (C*B*C**B** is cyclic, look at Antiparallels.html ).
- Triangles A*B**C** and GA'A" are (anti)similar, hence A'A* = A*A''. Analogously is proved that B'B*=B*B" and C'C*=C*C''.
- It follows A'A* = A*A'' = B'B* = B*B'' = C'C* = C*C''.
- Since I is on the medial line of A'A'', the distances IA', IA'' are equal and they are also equal with IB' and IB'' and IC', IC'', proving points A', A'', B', B'', C', C'' to be all on the same circle with center at I.

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