In the diagram, $D$ is on side $A C$ of $\triangle A B C$ so that $B D$ is perpendicular to $A C$. Also, $\angle B A C=60^{\circ}$ and $\angle B C A=45^{\circ}$. If the area of $\triangle A B C$ is $72+72 \sqrt{3}$, what is the length of $B D$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_56af4dba662dd63eccc5g-2.jpg?height=292&width=425&top_left_y=2231&top_left_x=1316)
Let $B D=h$.

Since $\angle B C A=45^{\circ}$ and $\triangle B D C$ is right-angled at $D$, then $\angle C B D=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}$.

This means that $\triangle B D C$ is isosceles with $C D=B D=h$.

Since $\angle B A C=60^{\circ}$ and $\triangle B A D$ is right-angled at $D$, then $\triangle B A D$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle. Therefore, $B D: D A=\sqrt{3}: 1$.

Since $B D=h$, then $D A=\frac{h}{\sqrt{3}}$.

Thus, $C A=C D+D A=h+\frac{h}{\sqrt{3}}=h\left(1+\frac{1}{\sqrt{3}}\right)=\frac{h(\sqrt{3}+1)}{\sqrt{3}}$.

We are told that the area of $\triangle A B C$ is $72+72 \sqrt{3}$.

Since $B D$ is perpendicular to $C A$, then the area of $\triangle A B C$ equals $\frac{1}{2} \cdot C A \cdot B D$. Thus,

$$
\begin{aligned}
\frac{1}{2} \cdot C A \cdot B D & =72+72 \sqrt{3} \\
\frac{1}{2} \cdot \frac{h(\sqrt{3}+1)}{\sqrt{3}} \cdot h & =72(1+\sqrt{3}) \\
\frac{h^{2}}{2 \sqrt{3}} & =72 \\
h^{2} & =144 \sqrt{3}
\end{aligned}
$$

and so $B D=h=12 \sqrt[4]{3}$ since $h>0$.

ANSWER: $12 \sqrt[4]{3}$