Morley’s Miracle (Proof)

Morley’s Miracle states that the trisectors of each angle of any triangle meet in three new points that form a new triangle that is always an equilateral triangle. 

Proof:

sin(⅓C) / c1 = sin(∠AEC) / b	(law of sines on ∆AEC)
∠AEC + ⅓A + ⅓C = 180°	(angle sum of ∆AEC)
∠AEC = 180° – (⅓A + ⅓C)	(subtract)
sin(⅓C) / c1 = sin(180° – (⅓A + ⅓C)) / b	(substitute)
sin(⅓C) / c1 = sin(⅓A + ⅓C) / b	(sin(180° – x) = sin x)
c1 = b sin(⅓C) / sin(⅓A + ⅓C)	(rearrange)
A + B + C = 180°	(angle sum of ∆ABC)
⅓A + ⅓B + ⅓C = 60°	(divide by 3)
⅓A + ⅓C = 60° – ⅓B	(subtract)
c1 = b sin(⅓C) / sin(60° – ⅓B)	(substitute)
c1 = 4b sin(⅓B) sin(⅓C) sin(60° + ⅓B) / (4 sin(⅓B) sin(60° – ⅓B) sin(60° + ⅓B))	(multiply num and denom by 4 sin (⅓B) sin(60° + ⅓B))
c1 = 4b sin(⅓B) sin(⅓C) sin(60° + ⅓B) / d	(let d be the denon)
d = 4 sin(⅓B) sin(60° – ⅓B) sin(60° + ⅓B)	(d value)
d = 4 sin(⅓B) (sin(60°)cos(⅓B) – cos(60°)sin(⅓B)) (sin(60°)cos(⅓B) + cos (60°)sin(⅓B))	(sine addition and subtraction)
d = 4 sin(⅓B) (√3/2cos(⅓B) – ½sin(⅓B)) (√3/2cos(⅓B) – ½sin(⅓B))	(sine and cosine values)
d = 4 sin(⅓B) (¾cos2(⅓B) – ¼sin2(⅓B))	(difference of squares)
d = 3 sin(⅓B) cos2(⅓B) – sin3(⅓B)	(distribute 4 sin(⅓B))
d = sin(⅓B) cos2(⅓B) – sin3(⅓B) + 2 sin(⅓B) cos2(⅓B)	(rearrange)
d = sin(⅓B)(cos2(⅓B) – sin2(⅓B)) + 2sin(⅓B)cos(⅓B)cos(⅓B)	(factor out sin(⅓B))
d = sin(⅓B)(cos(⅓B)cos(⅓B) – sin(⅓B)sin(⅓B)) + (sin(⅓B)cos(⅓B) + sin(⅓B)cos(⅓B))cos(⅓B))	(rearrange)
d = sin(⅓B)cos(⅓B + ⅓B) + sin(⅓B + ⅓B)cos(⅓B))	(sine addition)
d = sin(⅓B)cos(2(⅓B)) + sin(2(⅓B))cos(⅓B))	(addition)
d = sin(⅓B  + 2(⅓B))	(sine addition)
d = sin(3(⅓B))	(addition)
d = sin(B)	(multiply)
c1 = 4b sin(⅓B) sin(⅓C) sin(60° + ⅓B) / sin(B)	(substitute)
c1 = (4b/sin(B)) sin(⅓B) sin(⅓C) sin(60° + ⅓B)	(rearrange)
K = ½ ac sin B	(area of ∆ABC)
4bK = 2 abc sin B	(multiply by 4b)
4b/sin B = 2abc/K	(divide by K sin B)
c1 = (2abc/K) sin(⅓B) sin(⅓C) sin(60° + ⅓B)	(substitute)
b3 = (2acb/K) sin(⅓C) sin(⅓B) sin(60° + ⅓C).	(similar argument to c1)
Let P = (2abc/K) sin(⅓B) sin(⅓C)	(P value assignment)
Let Q = 60° + ⅓B	(Q value assignment)
Let R = 60° + ⅓C	(R value assignment)
c1 = P sin Q	(substitution)
b3 = P sin R	(substitution)
a22 = b32 + c12 – 2b3c1cos(⅓A)	(law of cosines on ∆AEF)
a22 = (P sin R)2 + (P sin Q)2 – 2(P sin R)(P sin Q)cos(⅓A)	(substitution)
a22 = P2 sin2R + P2 sin2Q – 2 P2 sin R sin Q cos (⅓A)	(square)
a22 = P2(sin2R + sin2Q – 2 sin R sin Q cos (⅓A))	(factor P2)
A + B + C = 180°	(angle sum of ∆ABC)
⅓A + ⅓B + ⅓C = 60°	(divide by 3)
⅓A + ⅓B + ⅓C + 60° + 60° = 60° + 60° + 60°	(add 60° and 60°)
⅓A + (60° + ⅓B) + (60° + ⅓C) = 180°	(rearrange)
⅓A + Q + R = 180°	(substitution)
⅓A = 180° – (Q + R)	(subtract (Q + R))
a22 = P2(sin2R + sin2Q – 2 sin R sin Q cos (180° – (Q + R)))	(substitution)
a22 = P2(sin2R + sin2Q + 2 sin R sin Q cos (Q + R))	(cos(180° – x) = -cos x)
a22 = P2(sin2R + sin2Q + 2 sin R sin Q (cos Q cos R – sin Q sin R))	(cosine addition)
a22 = P2(sin2R + sin2Q + 2 sin R sin Q cos Q cos R – 2 sin2R sin2Q)	(distribute)
a22 = P2(sin2R – sin2R sin2Q + sin2Q – sin2R sin2Q + 2 sin R sin Q cos Q cos R)	(rearrange)
a22 = P2(sin2R(1 – sin2Q) + sin2Q(1 – sin2R) + 2 sin R sin Q cos Q cos R)	(factor)
a22 = P2(sin2R cos2Q + sin2Q cos2R + 2 sin R sin Q cos Q cos R)	(1 – sin2Q = cos2Q)
a22 = P2(sin2R cos2Q + 2 sin R sin Q cos Q cos R + sin2Q cos2R)	(rearrange)
a22 = P2(sin R cos Q + sin Q cos R)2	(factor)
a22 = P2(sin(R + Q))2	(sine addition)
a22 = P2(sin(180° – (R + Q)))2	(sin(180° – x) = sin x)
a22 = P2(sin(⅓A))2	(substitution)
a2 = P sin(⅓A)	(square root)
a2 = ((2abc/K) sin(⅓B) sin(⅓C)) sin(⅓A)	(substitution)
a2 = (2abc/K) sin(⅓A) sin(⅓B) sin(⅓C)	(rearrange)
b2 = (2bca/K) sin(⅓B) sin(⅓C) sin(⅓A)	(similar argument to a2)
c2 = (2cab/K) sin(⅓C) sin(⅓A) sin(⅓B)	(similar argument to a2)
b2 = (2abc/K) sin(⅓A) sin(⅓B) sin(⅓C)	(rearrange)
c2 = (2abc/K) sin(⅓A) sin(⅓B) sin(⅓C)	(rearrange)
a2 = b2 = c2 = (2abc/K) sin(⅓A) sin(⅓B) sin(⅓C)	(substitution)
∆DEF is equilateral	(a2, b2, and c2 are congruent)

Q.E.D.