Riemann surface
matlab

We construct a function $f: D \rightarrow D^{\prime}$ such that $f(O)=O^{\prime}, f(A)=A^{\prime}$ and a point
From the axiom of segment and angle construction $\Rightarrow$ that the so constructed function is bijective, establishing a biunivocal correspondence between the elements of the two sets.
As:
$$\left.\begin{array}{l} |O P|=\left|O^{\prime} P^{\prime}\right| \ |O Q|=\left|O^{\prime} Q^{\prime}\right| \ \widehat{P O Q} \equiv P^{\top} O^{\prime} Q^{\prime} \end{array}\right} \Rightarrow \triangle O P Q \equiv P^{\prime} \widehat{O}^{\prime} Q^{\prime} \Rightarrow|P Q|=\left|P^{\prime} Q^{\prime}\right|,(\forall) P, Q \in D \Rightarrow D \equiv D^{\prime}$$

$f: M \rightarrow M^{\prime}$ is an isometry $\Rightarrow f$ is bijective and $(\forall) P, Q \in M$ we have $|P Q|=$ $|f(P), f(Q)|, f$-bijective $\Rightarrow f$ – invertible and $f^{-1}$-bijective.
$$\begin{gathered} \left|P^{\prime} Q^{\prime}\right|=|f(P) ; f(Q)|=|P Q| \ \left.\left|f^{-1}\left(P^{\prime}\right) ; f^{-1}\left(Q^{\prime}\right)\right|=\left|f^{-1}(f(P)), f^{-1}(f(Q))\right|=|P Q|\right) \Rightarrow \ \left|P^{\prime} Q^{\prime}\right|=\left|f^{-1}\left(f\left(P^{\prime}\right)\right), f^{-1}\left(f\left(Q^{\prime}\right)\right)\right|,(\forall) P^{\prime}, Q^{\prime} \in M, \end{gathered}$$
therefore $f^{-1}: M^{\prime} \rightarrow M$ is an isometry.

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