|Let's begin by looking at the sides GF, FH, and
of the given equilateral triangle FGH. We will create isosceles
triangles JGF, KFH, and IHG whose base angles 1,
and 3 satisfy the equation and inequalities: 1 + 2
+ 3 = 120o;
2 < 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
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
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
(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.
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