Construct an equilateral triangle given its altitude is $3.2 \, cm$. Provide justification for your construction.

(N/A) Steps of Construction:
$1.$ Draw a line $l$.
$2.$ Mark any point $D$ on the line $l$.
$3.$ At point $D$,draw a perpendicular line $\overline{DX} \perp l$ and mark a point $A$ on $\overline{DX}$ such that $DA = 3.2 \, cm$.
$4.$ At point $A$,construct rays $AB$ and $AC$ such that $\angle DAB = 30^{\circ}$ and $\angle DAC = 30^{\circ}$,meeting line $l$ at points $B$ and $C$ respectively.
$5.$ $\triangle ABC$ is the required equilateral triangle.
Justification:
In $\triangle ABD$,$\angle ADB = 90^{\circ}$ and $\angle DAB = 30^{\circ}$,so $\angle ABD = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
Similarly,in $\triangle ACD$,$\angle ADC = 90^{\circ}$ and $\angle DAC = 30^{\circ}$,so $\angle ACD = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
Also,$\angle BAC = \angle DAB + \angle DAC = 30^{\circ} + 30^{\circ} = 60^{\circ}$.
Since all angles are $60^{\circ}$,$\triangle ABC$ is an equilateral triangle.