16. Let $\triangle A B C$ be an acute triangle, $B C=5$. $E$ is a point on $A C$ such that $B E \perp A C$, $F$ is a point on $A B$ such that $A F=B F$. Moreover, $B E=C F=4$. Find the area of the triangle.
(2 marks)
16. $\triangle A B C$ is an acute triangle, $B C=5$. $E$ is a point on $A C$ such that $B E \perp A C$; $F$ is a point on $A B$ such that $A F=B F$; and $B E=C F=4$. Find the area of $\triangle A B C$.
16. $8 \sqrt{3}-6$
16. Let $D$ be the foot of the perpendicular from $F$ to $A C$. Note that $E C=\sqrt{B C^{2}-B E^{2}}=\sqrt{5^{2}-4^{2}}=3$. On the other hand, $F D=\frac{1}{2} B E=2$. Therefore
$$
D C=\sqrt{C F^{2}-F D^{2}}=\sqrt{4^{2}-2^{2}}=2 \sqrt{3} .
$$

It follows that $A D=D E=2 \sqrt{3}-3$. Thus
$$
\begin{aligned}
{[A B C] } & =[A B E]+[B E C] \\
& =\frac{1}{2}(4)[2(2 \sqrt{3}-3)]+\frac{1}{2}(4)(3) \\
& =8 \sqrt{3}-6
\end{aligned}
$$