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

(N/A) Steps of construction:
$(1)$ Let $AB$ be the given ray with initial point $A$. Extend $AB$ to the left to form a line $MAB$.
$(2)$ Taking $A$ as the center and any radius,draw an arc to intersect the line $MAB$ at points $X$ and $Y$.
$(3)$ Taking $X$ and $Y$ as centers and a radius greater than $\frac{1}{2} XY$,draw two arcs that intersect at point $P$ above the line $MAB$.
$(4)$ Draw ray $AC$ passing through $P$. Thus,$\angle CAB = 90^{\circ}$ is constructed.
$(5)$ Let the intersection of the arc (centered at $A$) and ray $AC$ be point $Z$.
$(6)$ Taking $Y$ and $Z$ as centers and a radius greater than $\frac{1}{2} YZ$,draw two arcs that intersect at point $Q$.
$(7)$ Draw ray $AQ$. Thus,$\angle QAB = 45^{\circ}$ is the required angle.
Justification:
Draw $PX$ and $PY$. In $\Delta PAX$ and $\Delta PAY$:
$AX = AY$ (Radii of the same arc)
$PX = PY$ (Radii of congruent arcs)
$PA = PA$ (Common side)
By $SSS$ congruence rule,$\Delta PAX \cong \Delta PAY$.
Therefore,$\angle PAX = \angle PAY$ $(CPCT)$.
Since $\angle PAX + \angle PAY = 180^{\circ}$ (Linear pair),$\angle PAY = \frac{180^{\circ}}{2} = 90^{\circ}$.
Thus,$\angle CAB = 90^{\circ}$.
Now,consider $\Delta AYQ$ and $\Delta AZQ$:
$AY = AZ$ (Radii of the same arc)
$YQ = ZQ$ (Radii of congruent arcs)
$AQ = AQ$ (Common side)
By $SSS$ congruence rule,$\Delta AYQ \cong \Delta AZQ$.
Therefore,$\angle QAY = \angle QAZ$ $(CPCT)$.
Since $\angle QAY + \angle QAZ = \angle ZAY = \angle CAB = 90^{\circ}$,
$\angle QAY = \frac{90^{\circ}}{2} = 45^{\circ}$.
Thus,$\angle QAB = 45^{\circ}$.