$\measuredangle C=120^\circ$ 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!

As $\angle BCD = 60^0, BD = CD \tan 60^0 = CD \sqrt3$
$CH = AC \cos 60^0 \implies AC = 2 \cdot CH$ as $\angle ACH = 60^0$
Area $\triangle ABC = \dfrac{1}{2} \cdot AC \cdot BD = \dfrac{1}{2} \cdot 2 \cdot CH \cdot CD \sqrt3 = 15 \sqrt3$
(as you already found that $CH \cdot CD = 15$)
