CEMC Fermat, 2024

We note first that $\triangle ACB$ has a right angle and a $60\degree$ angle and so it is a $30\degree$-$\,60\degree$-$\,90\degree$ triangle.

Since $AB = \sqrt{3}$, then using the
known ratios of side lengths, we can see that $BC = 1$ and $AC = 2$.

Next, we note that $\triangle ACE$
has two $45\degree$ angles and so is
an isosceles right-angled triangle.

This means that $CE = AC = 2$ and
$\angle ACE = 90\degree$.

Also, $AE = \sqrt{2}AC = 2\sqrt{2}$.

Further, since $\angle BCD$ is a
straight angle, then $$\angle ECD = 180\degree - \angle ACB - \angle ACE = 180\degree - 60\degree - 90\degree = 30\degree$$ Since $\triangle CED$ has a $30\degree$ angle and a right-angle, it is
also a $30\degree$-$\,60\degree$-$\,90\degree$ triangle.

Using the known ratios of sides, since $CE = 2$, we have $DE = 1$ and
$CD = \sqrt{3}$.

Therefore, the perimeter of $ABDE$ is
$$AB + BC + CD + DE + AE = \sqrt{3} + 1 + \sqrt{3} + 1 + 2\sqrt{2} = 2 + 2\sqrt{2} + 2\sqrt{3}$$