Morley's Theorem

Morley's Theorem

Morley's Theorem states that in any triangle, if we trisect each angle, then the three points of intersection of adjacent trisectors form an equilateral triangle.

To prove this theorem, we begin by drawing an arbitrary triangle ABC and trisecting each of its angles. Let the trisectors of angles A, B, and C intersect each other at points D, E, and F, respectively, as shown in the figure below.

[Insert a diagram of the trisected triangle ABC with points D, E, and F labeled.]

We will now show that the triangle DEF is equilateral. To do this, we will first prove two lemmas.

Lemma 1: The angles DFE, EFD, and FED are all equal.

Proof: Since the trisectors of angle A intersect at point D, we know that ∠BDF = ∠CDF. Similarly, since the trisectors of angle B intersect at point E, we know that ∠CAE = ∠ABE. Therefore, we have:

∠DFE = 180° - (∠BDF + ∠CAE) = 180° - (∠CDF + ∠ABE)

= ∠FDE + ∠EDF

Similarly, we can show that ∠DEF = ∠EFD + ∠FED and ∠EDF = ∠DFE + ∠FDE. Adding these three equations, we get:

2(∠DFE + ∠EFD + ∠FED) = 180°

Therefore, ∠DFE + ∠EFD + ∠FED = 60°, which implies that each of these angles is 60°.

Lemma 2: The segments AD, BE, and CF are concurrent.

Proof: To prove this, we will use Ceva's Theorem. Let X be the point where the line segment AD intersects the line segment BE, as shown in the figure below.

[Insert a diagram of triangle ABC with points D, E, and X labeled.]

Then, by Ceva's Theorem, we have:

(AX/EX) × (EB/BF) × (FC/CA) = 1

Since AX = DX and BF = CF, we can rewrite this equation as:

(DX/EX) × (EB/CF) × (FC/CA) = 1

But since the trisectors of angles A and B intersect at point D and E, respectively, we have:

DX/EX = BD/BE and EB/CF = AE/AF

Therefore, we can rewrite the equation as:

(BD/BE) × (AE/AF) × (FC/CA) = 1

Simplifying, we get:

BD/DC = BE/EA

This implies that the point X lies on the third trisector, namely the trisector of angle C.

Now, using Lemma 1, we can see that the triangle DEF is equilateral, since all of its angles are equal to 60°. Therefore, we have proven Morley's Theorem.

In summary, Morley's Theorem states that in any triangle, if we trisect each angle, then the three points of intersection of adjacent trisectors form an equilateral triangle. We have proven this theorem by first showing that the angles of the triangle DEF formed by the trisectors are all equal, and then using Ceva's Theorem to show that the segments AD, BE, and CF are concurrent, which implies that the triangle