Rename the unknown pentagon as $ABCDE$ and let $a,b,c,d,e$ be the external bisectors of its angles at $A,B,C,D,E,$ respectively. $\mathcal{A}(\ell)$ denotes the axial symmetry across $\ell.$ Oriented lines $AB,EA$ are homologous under the composition $ \mathcal{A} \equiv \mathcal{A}(b) \circ \mathcal{A}(c) \circ \mathcal{A}(d) \circ \mathcal{A}(e).$ But, $\mathcal{A}(b) \circ \mathcal{A}(c)$ is a rotation $\mathcal{R}_1$ with center $b \cap c$ and $\mathcal{A}(d) \circ \mathcal{A}(e)$ is another rotation $\mathcal{R}_2$ with center $d \cap e.$ Therefore, $AB,EA$ are homologous under certain rotation $\mathcal{R}_3 \equiv \mathcal{R}_1 \circ\mathcal{R}_2 .$ To define $\mathcal{R}_3,$ pick two arbitrary points $U,V$ in the plane and construct their images $U',V'$ under $\mathcal{A}.$ Then, perpendicular bisectors of $UU'$ and $VV'$ meet at the center $O$ of ${\mathcal{R}_3}$ and $\angle UOU'=\angle VOV'=\omega$ is its rotational angle. Consequently, $A$ is the orthogonal projection of $O$ onto $a$ and $\angle(a, AB)=\frac{_1}{^2}\omega.$ Once the point $A$ and the line passing through $A,B$ are constructed, the construction of the remaining vertices $B,C,D,E$ is straightforward.