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Explanation ### Zaslavsky's Theorem

Given triangle ABC, point P, and reflection A'B'C' of ABC in P. Let three parallel lines through A', B', and C' intersect BC, AC, and AB in X, Y, Z, respectively. Then X, Y, Z are collinear.

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The theorem was posted at the Yahoo's Hyacinthos discussion forum by Alexey Zaslavsky, and its solution that we follow below was posted by Darij Grinberg at the geometry-college newsgroup.

We are to show that three points on the side lines of a triangle are collinear. The situation calls for Menelaus' theorem. It suffices to show that

 (1) BX/CX × CY/AY × AZ/BZ = 1.

(Here and below all segments are directed.) Let the line through C parallel to the "parallel triplet" intersect BA' at N, AB' at K and A'B' at Z'.

First note that Z' is a reflection of Z in P, so that B'Z' = -BZ. Also, since the points A, B, and Z are collinear, so are their reflections A', B', and Z'. In particular, A'Z' = -AZ. Thus

 (2) B'Z'/A'Z' = BZ/AZ.

Further

 BX/CX × CY/AY × AZ/BZ = BA'/NA' × B'K/B'A × AZ/BZ = BA'/NA' × B'K/(-BA') × AZ/BZ = -B'K/NA' × AZ/BZ = B'Z'/A'Z' × AZ/BZ = BZ/AZ × AZ/BZ = 1.

Q.E.D.

Michel Cabart has observed that the proof can be shortened by a simple device:

Let us draw any oriented axis with origin P, not parallel (for example perpendicular) to the parallel triplet. All points have an ordinate tA, tB, etc. and we have tZ = -tC, tX = -tA, tY= -tB. Thus

 (XB/XC) × (YC/YA) × (ZA/ZB) = [(tB + tA)/(tB + tC)] × [(tC + tB)/(tA + tB)] × [(tA + tC)/(tA + tB)] = 1.

In a private correspondence Alexey Zaslavsky made the following remark:

 My original formulation was stronger: this line and the sidelines of the two triangles touch some conic. And there is nice proof of this fact. Let the line l touch the conic inscribed in both triangles and intersect the lines A'C' and B'C' in points P, Q. It is sufficient to prove that AQ and BP are parallel. This follows by applying Brianchon theorem to hexagon ABXQPY where X, Y are the infinite points of BC and AC. 