Morley's Theorem, a Proof
Brian Stonebridge (University of Bristol, UK)
Bill Millar
In Figure 1, the near trisectors of the internal angles at the vertices A, B, and C of a triangle meet in X, Y, and Z. Morley's theorem states that the triangle XYZ is equilateral. We give here a direct Euclidean proof.
Let the far trisectors meet in P, Q, and R. The following proof that PX, QY, and RZ are concurrent is purely Euclidean; a stronger result can be proved using projective geometry. Figure 1 shows the angle values α, β, γ, implying α + β + γ = π/3, and ∠QXR = π - β - γ. Since Y is the incentre of ΔAQC, it follows that
Assume PX, QY and RZ meet in pairs at U, V, and W. By transitivity, either U, V, W all coincide or are all distinct. We show that the assumption that the three points are distinct leads to a contradiction. Indeed, this might happen in two ways, as illustrated in Figures 2 and 3 (below.) We focus on the diagram of Figure 2, the other one being entirely analogous. Sum the angles of quadrilateral QURX, using the above results, to find
(1) | d ≠ 0. |
Choose X_{1} on QX such that UX_{1} is parallel to PX. Then the angles ZUQ, QUX_{1}, X_{1}UV all are π/3. Now choose X_{2} and X_{3} on PX such that
Denote X_{2}X by x, then
(2) | VX = UZ - d + x. |
Similarly, defining y and z in the same way as for x,
(3) | WY = VX - d + y, |
(4) | UZ = WY - d + z. |
Adding equations (2)-(4) gives
(5) | x + y + z = 3d. |
Now extend the line PX to H, such that ∠X_{2}X_{1}H = π/3, and hence
X_{2}H | = X_{1}X_{2} × √3 | |
= (d × √3) × √3 | ||
= 3d. |
Choose G on X_{2}H such that
Now complete the proof of Morley's Theorem. As was observed earlier, in Figure 2,
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