Inscribed circle in a trapezoid BA = 3. The radius of the circle is 2. What is DC?
I know that AG = AF = 1. And i know that FC = CE. And DC = DE + EC. I drew a line from point A to line DC to create a right triangle, but I can’t figure out where to go from there


Attach a coordinate system to the figure, and let its origin be at point $D$.
Then $Z = (2, 2)$, and $A = (3, 4) $, and $C = (x_1, 0) $
When want to solve for $x_1$ such that the distance between $Z$ and $AC$ is $2$.
The equation of $AC $ is $ y - 4 = \dfrac{-4}{x_1-3} (x - 3) $
We want to make sure the coefficients of $x$ and $y$ are negative.
Applying the distance formula using the above line equation and the point $Z = (2,2)$, we obtain
$ d = 2 = \dfrac{ -2 + 4 - \dfrac{4}{x_1 - 3} (2 - 3) }{ \
\sqrt{1 + \dfrac{16}{(x_1 - 3)^2}}} $
From which,
$ 4 \left( 1 + \dfrac{16}{(x_1 - 3) ^2 } \right ) = \left(2 + \dfrac{4}{x_1 - 3}\right)^2 = 4 + \dfrac{16}{(x_1 - 3)^2} + \dfrac{16}{(x_1 - 3)}  $
So that,
$\dfrac{ 3 }{ ( x_1 - 3) ^ 2 } = \dfrac{1}{ (x_1 - 3) } $
And therefore,
$ (x_1 - 3) = 3 $
Hence $x_1 = 6 $
And this is equal to the the length $DC$.
