Solution:

Consider a circle of positive radius in the plane and inscribe a regular heptagon $A B C D E F G$ in it. Since the seven vertices of this heptagon are coloured by three colours, some three vertices have the same colour, by pigeon-hole principle. Consider the triangle formed by these three vertices. Let us call the part of the circumference separated by any two consecutive vertices of the heptagon an arc. The three vertices of the same colour are separated by arcs of length $l, m, n$ as we move, say counter-clockwise, along the circle, starting from a fixed vertex among these three, where $l+m+n=7$. Since, the order of $l, m, n$ does not matter for a triangle, there are four possibilities: $1+1+5=7 ; 1+2+4=7 ; 1+3+3=7 ; 2+2+3=7$. In the first, third and fourth cases, we have isosceles triangles. In the second case, we have a triangle whose angles are in geometric progression. The four corresponding figures are shown below.

(i)

(ii)

(iii)

(iv)

In (i), $A B=B C$; in (iii), $A E=B E$; in (iv), $A C=C E$; and in (ii) we see that $\angle D=\pi / 7$, $\angle A=2 \pi / 7$ and $\angle B=4 \pi / 7$ which are in geometric progression.