Construct an angle of $45^{\circ}$ at the initial point of a given ray and justify the construction.

(N/A) Steps of construction:
$I.$ Draw a ray $\overrightarrow{OA}$.
$II.$ Taking $O$ as the centre and with a suitable radius,draw a semicircle such that it intersects $\overrightarrow{OA}$ at $B$.
$III.$ Taking $B$ as the centre and keeping the same radius,cut the semicircle at $C$. Similarly,cut the semicircle at $D$ and $E$,such that $\widehat{BC} = \widehat{CD} = \widehat{DE}$.
$IV.$ Bisect $\angle BOC$ to get a ray $\overrightarrow{OF}$ such that $\angle BOF = 30^{\circ}$.
$V.$ Bisect $\angle BOF$ to get a ray $\overrightarrow{OG}$ such that $\angle BOG = 45^{\circ}$.
Justification:
Since $\widehat{BC} = \widehat{CD} = \widehat{DE}$,we have $\angle BOC = \angle COD = \angle DOE = 60^{\circ}$ (as they subtend equal arcs at the centre).
Since $\overrightarrow{OF}$ is the bisector of $\angle BOC$,$\angle BOF = \frac{1}{2} \times 60^{\circ} = 30^{\circ}$.
Since $\overrightarrow{OG}$ is the bisector of $\angle FOC$ (where $\angle FOC = 30^{\circ}$),$\angle FOG = \frac{1}{2} \times 30^{\circ} = 15^{\circ}$.
Therefore,$\angle BOG = \angle BOF + \angle FOG = 30^{\circ} + 15^{\circ} = 45^{\circ}$.