Thales' Triangle

As $CD=\dfrac{BC}{3}$ , we have $AC+\dfrac{BC}{3}=12 cm$

Now, $\frac{CD}{BC}=\frac{1}{3} \\\implies \frac{CD}{BC-CD}=\frac{1}{3-1} \\\implies \frac{CD}{BD}=\frac{1}{2}\\\implies \frac{AC}{AB}= \frac{1}{2}$ The last equality follows by an application of angle bisector theorem .

$\text{Perimeter}= AB+BC+CA=2\left(AC\right)+BC+AC\\=3\left(AC+\frac{BC}{3}\right)=3\left(12 cm\right)=\boxed{36 cm}$

I'll be using standard conventions for $\Delta$ s in my solution.Let $\dfrac{A}{2}=X$ From the given data we see $\dfrac{a}{3}+b=12\implies a+3b=36$ . Also applying $\sin$ rule for $\Delta$ $ADC$ we have $\dfrac{\sin(X)}{\dfrac{b}{3}}=\dfrac{\sin(C)}{AD}$ .Now applying $\sin$ rule for $\Delta$ $ADB$ we have $\dfrac{\sin(X)}{\dfrac{2b}{3}}=\dfrac{\sin(B)}{AD}$ . Now dividing one equation by the other we have $\sin(C)=2\sin(B) \implies c=2b$ . The perimeter of $\Delta$ $ABC$ $\color{#3D99F6}{\beta}=a+b+c=a+b+2b=\color{#D61F06}{\boxed{36}}$