In his book ALGEBRA, $2^{\text{nd}}$ ed., Artin proves (Ch. 6 - Symmetry, Lemma 6.3.5) that an orientation preserving isometry $f$ that has the form $m=t_a \rho_\theta$ with $\theta \neq0$, is a rotation through the angle $\theta$ about a point in the plane. He does so by proving the existence of a point $p$ in the plane about which $f$ is a rotation; $p=(I-\rho_\theta)^{-1}a$ . So far, so good.
Now, he gives a geometrical construction that is solely based on Book I of Euclid's elements, to find $p$, given the vector $a$ and the angle $\theta$.
The point $p$ is the fixed point of the isometry $t_a \rho_\theta$, and it can be found geometrically, as illustrated below. The line $l$ passes through the origin and is perpendicular to the vector $a$ (Euc. I.12). The sector with angle $\theta$ is situated so as to be bisected by $l$ (Euc. I.8, I.9) and the fixed point $p$ is determined by inserting the vector $a$ into the sector, as shown.
Now if I took an arbitrary point $P$, rotated it anticlockwise by $\theta$ with respect to $O$ to get $P'$ and then translated $P'$ by the vector $a$ to get $P''$ ; instead if I rotated $P$ w.r.t. $p$ to get $Q$, then $P''$ and $Q$ would coincide. This is indeed true by theorem (and I verified with actual construction too.) However, I think it (this last result) can be proved by Euclidean system of axioms and postulates too, but I can't figure out more than that the triangles $PpQ$, $POP'$ and $pOq$ are all similar.