Here the statement of the theorem from Mathworld:

Theorem. If isosceles triangles with apex angles $$2 k \pi/n$$ are erected on the sides of an arbitrary $$n$$-gon $$A_0$$, and if this process is repeated with the $$n$$-gon $$A_1$$ formed by the free apices of the triangles, but with a different value of $$k$$, and so on until all values $$1 \le k \le n-2$$ have been used in arbitrary order, then a regular $$n$$-gon $$A_{n-2}$$ is formed whose centroid coincides with the centroid of $$A_0$$.

Proof. You can represent an $$n$$-gon by the list of its vertices thought of as complex numbers, that is, by a vector in $$\mathbb{C}^n$$. Take an $$n$$-gon $$A$$ and let’s try to find a formula for the $$n$$-gon $$B$$ formed by the free vertices of similar triangles built on the sides of $$A$$. For a fixed complex number $$\alpha$$ determining the shape of the similar triangles, we have $$\alpha (A_i - B_i) = A_{i+1} - B_i$$, that is, $$B_i = (1-\alpha)^{-1}(A_{i+1} - \alpha A_i).$$

In terms of the linear operator $$S : \mathbb{C}^n \to \mathbb{C}^n$$ that cyclically permutes the coordinates one place, we have $$B = (1-\alpha)^{-1}(S-\alpha I) A$$, where $$I$$ is the identity. Taking centroids is a linear functional on this space of $$n$$-gons and is invariant under $$S$$, so we easily see that $$B$$ has the same centroid as $$A$$, which will prove the final remark in the theorem once we’ve proved the rest.

This means that the polygon $$A_{n-2}$$ that we need to show is regular is obtained from $$A_0$$ by applying the composition of the operators $$(1-\omega^k)^{-1} (S-\omega^k I)$$ for $$k=1,2,\ldots,n-2$$, where $$\omega = \exp(2 \pi i/n)$$. (These commute because they are all polynomials in the same operator $$S$$.)

To proceed we need a criterion for when a polygon is regular. Well, $$P$$ is a regular $$n$$-gon if each side is obtained from the next by rotating an angle of $$2\pi/n$$, that is, if $$P_{i+1}-P_i = \omega (P_{i+2} - P_{i+1})$$. In terms of $$S$$, $$P$$ is regular if it is in the kernel of $$(S-I)(I-\omega S)$$, or better yet, of $$(S-I)(S-\omega^{n-1}I)$$.

So to prove the theorem we only need to show that the composition of all the $$S - \omega^k I$$ for $$k=0,1,\ldots,n-1$$ is zero, which is true since that composition is just (a factored form of) $$S^n - I = 0$$.    □