The Equilateral Triangle Wizardry

@Maharnab Mitra Lami's theorem tells us that $3:4:5=\sin\angle BMC : \sin \angle CMA : \sin\angle AMB$ , but how to prove that $3:4:5=\angle BMC : \angle CMA : \angle AMB$ ?

It, of course, isn't generally true that $\sin A : \sin B = A : B$ . Is this where the fact that the triangle is equilateral is used (we haven't used the fact that the triangle is equilateral here yet, Lami's theorem doesn't require it).

Could anyone prove this here?

I have to publish here my solution for it's gone "private"...without remedy.

For simplicity let’s name AM=a, BM=b, CM=c and

$\angle AMB=\alpha, \angle BMC=\beta, \angle CMA=\gamma$

Applying cosine rule, we can now write

$1)\space\space\space a^2+b^2-2ab\cos \alpha=b^2+c^2-2bc\cos\beta$

$2)\space\space\space a^2+b^2-2ab\cos \alpha =a^2+c^2-2ac\cos\gamma$

From 1) we then get the value of $\beta$ in function of $\alpha$ , and from 2) the value of $\gamma$ in function of $\alpha$ , implicitly:

$3)\space\space\space \cos \beta=\frac{c^2-a^2+2ab \cos\alpha}{2bc}$

$4)\space\space\space \cos \gamma=\frac{c^2-b^2+2ab \cos\alpha}{2ac}$

Plugging in the values of a, b, c (3k, 4k, 5k) we hence have:

$5)\space\space\space \cos \beta=\frac{2+3 \cos\alpha}{5}$

$6)\space\space\space \cos \gamma=\frac{3+8 \cos\alpha}{10}$

Let’s remember, at this point that:

$\alpha +\beta +\gamma=2 \pi$

hence we can write

$\alpha=-\beta -\gamma$

and taking the cosine of both sides

$\cos \alpha=\cos(\beta+\gamma)$

and, at last

* $7)\space\space\space \cos \alpha=\cos\beta\cos\gamma-\sqrt{1-\cos^2 \beta} \sqrt{1-\cos^2 \gamma}$ *

which, through the equations 5) and 6) and some iterations, quickly give us the desired answer.