# Working Backwards

 Let's begin by looking at the sides GF, FH, and HG of the given equilateral triangle FGH.  We will create isosceles triangles JGF, KFH, and IHG whose base angles 1, 2, and 3 satisfy the equation and inequalities: 1 + 2 + 3 = 120o; 1 < 60o,     2 < 60o, 3 < 60o. If we extend the sides of the isosceles triangles below their bases to points A, B, and C, we can infer the measurements of some of the other angles since we know 1 + 2 + 3 + 60o = 180o.  Please note these new angles in the diagram below. In order to continue, we must use a lemma based from the ideas of the incenter of a triangle.  We will call the incenter of a triangle ABC, L.  The lemma states that we can think of L as lying on the bisector of angle A in our triangle ABC at such a distance that angle BLC = 90o + (0.5)A.  Using this idea, and applying it to point H in the triangle JBC, we can observe that the line HJ bisects the angle at J.  Also, the half angle at J is given by (90o - 1), and angle BHC = (180o - 1) = 90o + (90o - 1).  Hence H is the incenter of the triangle JBC.  In a similar manner we can argue that G is the incenter of KAC, and F of IAB.  Therefore all three small triangles at C are equal; likewise at A and B.  In other words, the angles of the triangle ABC are trisected. Now, the three small angles at A are each (1/3)A = (60o - 1); similarly at B and C.  Thus: 1 = (60o - (1/3)A), 2 = (60o - (1/3)B), and 3= (60o - (1/3)C). Finally, by choosing these values for the base angles of our isosceles triangles, we can ensure that the above procedure yields a triangle ABC that is similar to any given triangle. QED

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