The diameter $AB$ of a circle of radius $3$ is extended to $D$ outside the circle so $BD=2$. Point $E$ satisfies $ED=6$, $ED\perp AD$. Segment $AE$ meets the circle at $C$ between $A,E$. What is the area of $\triangle ABC$?

1. $AD=2r+k$ ($r=3,k=2$: $AD=8$), $AE=\sqrt{AD^2+ED^2}=\sqrt{64+36}=10$.
2. $[AED]=\frac12(8)(6)=24$; $[ABC]=\left(\frac{2r}{AE}\right)^2[AED]=\left(\frac6{10}\right)^2(24)=\frac{216}{25}$.
3. The answer is $\boxed{\dfrac{216}{25}}$.