Construct an equilateral triangle if its altitude is $6\, cm$. Give justification for your construction.

(N/A) Steps of construction:
$1$. Draw a line $XY$.
$2$. Take any point $D$ on the line $XY$.
$3$. Construct a perpendicular $PD$ to $XY$ at point $D$.
$4$. Cut a line segment $AD = 6\, cm$ from $PD$.
$5$. At point $A$,construct angles $\angle CAD = 30^{\circ}$ and $\angle BAD = 30^{\circ}$ such that $B$ and $C$ lie on the line $XY$.
$6$. Join $AB$ and $AC$. Thus,$\triangle ABC$ is the required equilateral triangle.
Justification:
In $\triangle ABC$,$AD \perp BC$. Since $\angle BAD = 30^{\circ}$ and $\angle CAD = 30^{\circ}$,then $\angle A = \angle BAD + \angle CAD = 30^{\circ} + 30^{\circ} = 60^{\circ}$.
Since $AD$ is the altitude,$\angle ADB = 90^{\circ}$. In $\triangle ABD$,$\angle B = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
Similarly,in $\triangle ACD$,$\angle C = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
Since all angles are $60^{\circ}$,$\triangle ABC$ is an equilateral triangle with altitude $AD = 6\, cm$.