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. |
|
|
|
|
|
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
|
Geoff's Page | Morley Main Page | Morley Proof Page