Proof by B. Stonebridge and B. Millar

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.

Morley's theorem

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 ∠ZQY = ∠XQY = β + π/6. The corresponding angles at R and P are γ + π/6, α + π/6.

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 ∠WUV = π/3. Hence ΔWUV is equilateral, with sides 2d, say. Thus we assume that

(1)

d ≠ 0.

Choose X₁ on QX such that UX₁ is parallel to PX. Then the angles ZUQ, QUX₁, X₁UV all are π/3. Now choose X₂ and X₃ on PX such that ∠PX₂X₁ = π/2, and X₁X₃ is parallel to UV. These constructions imply ∠X₂X₁X = β (because in ΔQWX, ∠WQX = β + π/6 and ∠QWX = π/3 hence ∠QXW = π/2 - β) and X₂X₃ = d - from the choice of X₃.

Morley's theorem

Denote X₂X by x, then VX = VX₃ - d + x. But VX₃ = UX₁ = UZ, from the congruence of triangles ZUQ and X₁UQ. (These triangles are congruent since they have a common side UQ, the angles at Q are β + π/6, and the angles at U are π/3.) Hence

(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 ∠X₂X₁H = π/3, and hence X₂H = 3d. Indeed, in an equilateral triangle the ratio of the altitude to a half-side equals √3. In ΔUVW, this means that X₁X₂ = d × √3. In ΔX₁X₂H (which is a half of an equilateral triangle), we have X₂H = X₁X₂ × √3. Multiplying through gives

	X₂H	= X₁X₂ × √3
		= (d × √3) × √3
		= 3d.

Choose G on X₂H such that ∠XX₁G = γ. Then ∠GX₁H = α. Finally, choose F on X₁G such that XF is normal to XX₁. Then ΔXX₁F and ΔY₂Y₁Y are similar. Since XX₁ > X₂X₁ = Y₂Y₁, then XF > y. Because ∠XFG = π/2 + γ, it is the greatest angle in ΔXFG. Hence, by Euclid (I.19), XG is the longest side, so y < XG. Similarly, z < GH. Including x = X₂X, it follows that x + y + z < 3d. This contradicts (5) and so the assumption d ≠ 0 was false. That is, PX, QY, and RZ are concurrent, with mutual angles of π/3.

Now complete the proof of Morley's Theorem. As was observed earlier, in Figure 2, UZ = UX₁, so that ΔZUX₁ is isosceles. Hence YQ, being the bisector at vertex U, is normal to the base ZX₁. But X₁ and X now the same point, so ZX is normal to YQ. Similarly, XY, YZ are normal to RZ, PX, implying that the ΔXYZ is equilateral.

Morley's theorem