Morley's Theorem

In a triangle $ABC$, draw the trisectors of the angles $A$, $B$, $C$. These trisectors intersect at $A'$, $B'$ and $C'$ as in the figure below. Prove that the triangle $A'B'C'$ is an equilateral triangle.

Proof: First, let $$a=\frac{\angle A}{3}, ~~b=\frac{\angle B}{3}, ~~c=\frac{\angle C}{3},$$ then $$a+b+c = 60^{o} .$$ Next, let $$a' = a + 60^{o}, ~b' = b + 60^{o}, ~c' = c + 60^{o},$$ then $$a + b' + c' = b + c' + a' = c + a' + b' = 180^{o}.$$

Now instead of starting with the triangle $ABC$, we start with an equilateral triangle $A'B'C'$. Next, outside the triangle $A'B'C'$, we draw three triangles $A'B'C$, $B'C'A$, and $C'A'B$ such that
$$\angle A'CB' = c,  ~~\angle A'B'C= a',  ~~\angle B'A'C = b',$$
$$\angle B'AC' = a,  ~~\angle B'C'A = b',  ~~\angle C'B'A = c',$$
$$\angle C'BA' = b,  ~~\angle C'A'B = c',  ~~\angle A'C'B = a' .$$
The reason we can do this is because $a + b ' + c ' = b + c ' +  a ' = c + a ' + b ' = 180^{o}$ as we have shown above.

Our job now is to prove
$$\angle BAC' = \angle CAB' = a, ~~ \angle ABC' = \angle CBA' = b, ~~ \angle BCA' = \angle ACB' = c .$$

To prove this, we draw $A'P$ and $A'Q$ such that  $$\angle C'A'P = 60^{o}, ~~\angle PA'B = c, ~~\angle B'A'Q = 60^{o}, ~~\angle QA'C = b.$$ Clearly, the two triangles  $A'C'P$ and $A'B'Q$ are equal and so we have $$A'P = A'Q.$$

Consider the two triangles $BPA'$ and $A'QC$. These two triangles have two pairs of equal angles so they are similar triangles. Thus, $$\frac{BP}{A'Q} = \frac{BA'}{A'C}.$$ Replacing $A'Q=PA'$ we get $$\frac{BP}{PA'} = \frac{BA'}{A'C}.$$ Combining this fact with $$\angle BPA' = \angle BA'C = 180^{o} - (b+c)$$ we deduce that the two triangles $BPA'$ and $BA'C$ are similar triangles.

It follows that $$\angle A'BC = \angle PBA' = b, ~~~\angle A'CB = \angle PA'B = c.$$

Prove similarly, we can obtain $$\angle BAC' = a, ~~~\angle ABC' =  b$$ and $$\angle CAB' = a, ~~~\angle ACB' = c,$$ and thus the theorem is proved.

No comments:

Post a Comment