This theorem was first discovered by the British mathematician Molly in 1904. Mill once mentioned this theorem to his classmates at Cambridge University, and later called it Mill Theorem. Although Mill discovered this theorem a long time ago, he never published it, and it took 20 years before it was officially published in Japan. In these 20 years, other mathematicians have also discovered this theorem.

I'll write about the process of construction method proof:

In any triangle ABC, I set X = 60+A/3, Y = 60+B/3 and Z = 60+C/3.

Then x+y+z+ 120=360.

Okay, now we're going to build it. Let's draw an equilateral triangle A/KOOC-0/B/KOOC-0/C/KOOC-0/,and then construct three triangles A 'B/KOOC-0/C/KOOC-0/A66 according to the following steps.

On the edge of B 1C 1, make two rays with an included angle of x to the end of A 1, and these two rays should be symmetrical about the vertical line of B 1C 1. Similarly, the edge of A 1C 1 also makes two rays with an included angle of y, and the edge of A 1B 1 also makes two rays with an included angle of z. Then, the intersections of these rays are A'B'C' and a triangle a' b1c respectively. God, how can I make it clear! ]

After a while, we only need to prove that triangle A'B'C' is similar to triangle ABC, and two straight lines on each corner of triangle A'B'C' are bisectors of angles.

Ok, now the angles B'A 1C 1=z, B'C 1A 1=x, and A 1B'C 1 equal to180.

In the same way, I wrote it all at once: three angles of triangle A'B'C' B 1A 'c 1, A 1B 'c 1, A 1C 'B 1 are equal to A.

Extend the intersection A2 of B'C 1 and C'B 1, then A2B 1C 1 is an isosceles triangle, and A2A 1 is the bisector of A2 (that is, B'A2C').

And x+60= angle B'C 1B 1=90+A2/2 (in this step, make a vertical line with an isosceles triangle and work it out by yourself).

Don't forget the theorem. If a point O in a triangle has an angle BOC=90+A/2, and AO bisects the angle A, then O is the center ABC of the triangle.

Then A 1 is the heart of triangle A2B'C'

Similarly, B 1 and C 1 are the hearts of B2C'A' and C2A'B'

And the angle b' a' c1= c1a' b' = b1a' c' = a/3, that is, A'=A and b' = b, c' = C.

So I got the similarity of ABC, A'B'C!

I'm exhausted. I didn't make it clear.

It is estimated that it needs to be revised again.

Other proof methods should use trigonometric functions.

There are too many ways to prove with trigonometric functions, and here is only one:

[intersection: AE CE meets E.

BD CD cross d

AF BF cross F]

Note that A=3α, B=3β, C=3γ, AE=m, AF=n, and the three sides of △ABC are A, B and C respectively.

Since 3α+3β+3γ = 180, α+β+γ = 60. α+β = 60-γ.

And nsin(α+β)=csinβ, so n = csinβ/sin (α+β) = csinβ/sin (60-γ).

Similarly, m=bsinγ/sin(60-β)

In △ABC, there is bsin3γ=csin3β, so

m/n =(sin 3β* sinγ* sin(60-γ))/(sin 3γ* sinβ* sin(60-β))

=(sin(60+β))/(sin(60+γ))

Because α+β+γ = 60, there are triangles with internal angles of 60+β, 60+γ and α, and the ratio of the two sides of α angle is.

(sin(60+β))/(sin(60+γ))=m/n

△EAF is similar to this triangle, so that

∠AFE=60 +β

∠AEF=60 +γ

The same method can prove that ∠ BFD = 60+α, and

∠AFB= 180 -(α+β)

therefore

∠EFA+∠AFB+∠BFD =(60+β)+( 180-α-β)+(60+α)= 300

So ∠ dfe = 60.

Similarly, the other two internal angles of △DEF are also 60.

So △DEF is an equilateral triangle.