Construct an angle of $90^{\circ}$ at the initial point of a given ray and justify the construction.

(N/A) Steps of construction:
$I.$ Draw a ray $OA$.
$II.$ Taking $O$ as the centre and a suitable radius,draw a semicircle,which cuts $OA$ at $B$.
$III.$ Keeping the radius the same,divide the semicircle into three equal parts such that $\widehat{BC} = \widehat{CD} = \widehat{DE}$.
$IV.$ Draw rays $\overrightarrow{OC}$ and $\overrightarrow{OD}$.
$V.$ Draw $\overrightarrow{OF}$,the bisector of $\angle COD$.
Thus,$\angle AOF = 90^{\circ}$.
Justification:
Since $O$ is the centre of the semicircle and it is divided into $3$ equal parts,
$\therefore \widehat{BC} = \widehat{CD} = \widehat{DE}$.
$\Rightarrow \angle BOC = \angle COD = \angle DOE$.
Since equal arcs subtend equal angles at the centre,and $\angle BOC + \angle COD + \angle DOE = 180^{\circ}$,we have $3 \angle BOC = 180^{\circ}$.
$\therefore \angle BOC = 60^{\circ}$.
Similarly,$\angle COD = 60^{\circ}$ and $\angle DOE = 60^{\circ}$.
Since $OF$ is the bisector of $\angle COD$,
$\therefore \angle COF = \frac{1}{2} \angle COD = \frac{1}{2}(60^{\circ}) = 30^{\circ}$.
Now,$\angle BOF = \angle BOC + \angle COF = 60^{\circ} + 30^{\circ} = 90^{\circ}$.
Thus,$\angle AOF = 90^{\circ}$.