Length of the altitude to the hypotenuse

Ewa, Alek and Ania solved the following task:

In a right-angled triangle $ABC$, the measure of the angle $ABC$ is $30^{\circ}$. The area of the triangle is $2\sqrt3$. Find the length of the altitude to the hypotenuse.

They all started by sketching a picture

and denoted $a=|BC|$, $b=|AC|$, $c=|AB|$, and $v=|CD|$.

Ewa knew that the altitude to the hypotenuse divides the triangle into two triangles that are similar to each other: $$ \Delta ACD\sim \Delta CBD $$ Using the similarity of triangles, she continued: $$ \begin{gather} \frac{v}{a}=\frac{b}{v }\cr v^2=ab \end{gather} $$ Since the area of the triangle $$ P=\frac12 ab=2\sqrt3 $$ she substituted $4\sqrt3$ for $ab$ into the formula for $v^2$ and got: $$ \begin{gather} v^2=4\sqrt3 \cr v=2\sqrt[4]{3} \end{gather} $$

Alek used the function sine: $$ \begin{gather} \sin ⁡30^{\circ}=\frac{b}{c} \cr b=\frac12 c \end{gather} $$ Then he expressed the area of the triangle $$ P=\frac12 bc \sin ⁡60^{\circ} $$ and got $$ \begin{gather} 2 \sqrt3= \frac12 \left(\frac12 c\right)c \frac{\sqrt3}{2} \cr 2\sqrt3=\frac{\sqrt3}{8} c^2 \cr c^2=16 \cr c=4 \end{gather} $$ Finally, he expressed the area of the triangle in terms of $c$ and $v$: $$ 2\sqrt3=\frac12 cv $$ From there he calculated: $$ v=\sqrt3 $$

Ania used the function tangent: $$ \begin{gather} \mathrm{tg}\,30^{\circ}=\frac{a}{b}\cr a=\frac{b}{\sqrt3} \end{gather} $$ Then she substituted for $a$ into the formula for the area of the triangle and got the lengths of $b$: $$ \begin{gather} P=\frac12 ab \cr 2\sqrt3=\frac12 \frac{b^2}{\sqrt3} \cr b^2=12 \cr b=2\sqrt3 \end{gather} $$

Finally, she determined the length of $v$ using the function cosine: $$ \begin{gather} \cos⁡\,60^{\circ}=\frac{v}{b} \cr v=b \cos⁡\,60^{\circ} \cr v=\sqrt3 \end{gather} $$ Which one of them proceeded correctly in solving the task?