A circle of radius $1$ is [tangent](https://artofproblemsolving.com/wiki/index.php/Tangent) to a circle of radius $2$. The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $\overline{AB}$  and $\overline{AC}$ are congruent. What is the area of $\triangle ABC$?
[asy] size(200); pathpen = linewidth(0.7); pointpen = black; real t=2^0.5; D((0,0)--(4*t,0)--(2*t,8)--cycle); D(CR((2*t,2),2)); D(CR((2*t,5),1)); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); D((2*t,2)--(2*t,4)); D((2*t,5)--(2*t,6)); MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W);[/asy]
$\textbf{(A) } \frac{35}{2}\qquad\textbf{(B) } 15\sqrt{2}\qquad\textbf{(C) } \frac{64}{3}\qquad\textbf{(D) } 16\sqrt{2}\qquad\textbf{(E) } 24\qquad$
Let the centers of the smaller and larger circles be $O_1$ and $O_2$ , respectively.
Let their tangent points to $\triangle ABC$ be $D$ and $E$, respectively.
We can then draw the following diagram:
[asy] size(200); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); real t=2^0.5; D((0,0)--(4*t,0)--(2*t,8)--cycle);  D(CR(D((2*t,2)),2)); D(CR(D((2*t,5)),1)); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0)); D((2*t,0)--(2*t,8)); D((2*t,2)--D(A)); D((2*t,5)--D(B)); D(rightanglemark((2*t,2),A,(4*t,0))); D(rightanglemark((2*t,5),B,(4*t,0))); MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); [/asy]
We see that $\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC$. Using the first pair of [similar triangles](https://artofproblemsolving.com/wiki/index.php/Similar_triangles), we write the proportion:

$\frac{AO_1}{AO_2} = \frac{DO_1}{EO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3$
By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), we have $AD = \sqrt{3^2-1^2} = \sqrt{8}$.
Now using $\triangle ADO_1 \sim \triangle AFC$,

$\frac{AD}{AF} = \frac{DO_1}{FC} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{FC} \Longrightarrow FC = 2\sqrt{2}$
Hence, the area of the triangle is \[\frac{1}{2}\cdot AF \cdot BC = \frac{1}{2}\cdot AF \cdot (2\cdot CF) = AF \cdot CF = 8\left(2\sqrt{2}\right) = \boxed{\textbf{(D) } 16\sqrt{2}}\]