$\measuredangle C=120^\circ$ and two altitudes $AH$ and $BD$ are altitudes of $\triangle ABC$ and $\measuredangle ACB=120^\circ$. If $S_{\triangle HCD}=\dfrac{15\sqrt3}{4},$ find the area of $\triangle ABC$.

$$S_{\triangle HCD}=\dfrac12\cdot CH\cdot CD\cdot\sin\measuredangle HCD=\dfrac{\sqrt3}{4}CH\cdot CD=\dfrac{15\sqrt{3}}{4}\ \implies CH\cdot CD=15$$
On the other hand $$S_{\triangle ABC}=\dfrac12\cdot AC\cdot BC\cdot\sin\measuredangle ACB=\dfrac{\sqrt3}{4}AC\cdot BC=?$$
I noted that $ABDH$ is inscribed, because $\measuredangle ADB=\measuredangle AHB=90^\circ$, so $$AC\cdot CD=BC\cdot CH.$$ I am stuck here. Thank you in advance!

Observe,
$$\angle ACH=\angle BCD=60^{\circ}\implies \angle CAH=CBD=30^{\circ}$$
In right-triangle $ AHC$,
$$\sin \angle CAH=\frac{CH}{AC}\implies AC=2\cdot CH $$
In right-triangle $ BDC$,
$$\sin \angle CBD=\frac{CD}{BC}\implies BC=2\cdot CD $$
Since you have arrived at $CH\cdot CD=15$,
$$\begin{align*}
\text{area}(\triangle ABC)&=\frac{1}{2}\cdot AC\cdot BC\cdot \sin \angle ABC \\
&=\frac{1}{2}\cdot (2\cdot CH)\cdot (2\cdot CD)\cdot \sin 120^{\circ}\\
&=\boxed{15\sqrt{3}}
\end{align*}$$
