In $\triangle{ADE}$ points $B$ and $C$ are on side $AD$ and points $F$ and $G$ are on side $AE$ so that $BG \parallel CF \parallel DE$, as shown. The area of $\triangle{ABG}$ is $36$, the area of trapezoid $CFED$ is $144$, and $AB = CD$. Find the area of trapezoid $BGFC$.

[center][img]https://snag.gy/SIuOLB.jpg[/img][/center]
1. Let \( AB = CD = x \) and \( BC = y \).

2. Given that the area of \(\triangle ABG\) is 36, we can express this area in terms of the area of \(\triangle ADE\). Since \(BG \parallel CF \parallel DE\), the triangles \(\triangle ABG\) and \(\triangle ADE\) are similar. The ratio of their areas is the square of the ratio of their corresponding sides:
   \[
   [ABG] = \left(\frac{x}{2x+y}\right)^2 \cdot [ADE]
   \]
   Given \([ABG] = 36\), we have:
   \[
   36 = \frac{x^2}{(2x+y)^2} \cdot [ADE]
   \]

3. Given that the area of trapezoid \(CFED\) is 144, we can express this area in terms of the area of \(\triangle ADE\). The area of trapezoid \(CFED\) is the difference between the area of \(\triangle ADE\) and the area of \(\triangle ACF\):
   \[
   [CFED] = [ADE] - [ACF]
   \]
   Since \(\triangle ACF\) is similar to \(\triangle ADE\), we have:
   \[
   [ACF] = \left(\frac{x+y}{2x+y}\right)^2 \cdot [ADE]
   \]
   Therefore,
   \[
   [CFED] = [ADE] - \left(\frac{x+y}{2x+y}\right)^2 \cdot [ADE] = \left(1 - \left(\frac{x+y}{2x+y}\right)^2\right) \cdot [ADE]
   \]
   Given \([CFED] = 144\), we have:
   \[
   144 = \left(1 - \left(\frac{x+y}{2x+y}\right)^2\right) \cdot [ADE]
   \]

4. Dividing the equation for \([CFED]\) by the equation for \([ABG]\), we get:
   \[
   \frac{144}{36} = \frac{\left(1 - \left(\frac{x+y}{2x+y}\right)^2\right) \cdot [ADE]}{\frac{x^2}{(2x+y)^2} \cdot [ADE]}
   \]
   Simplifying, we get:
   \[
   4 = \frac{(2x+y)^2 \left(1 - \left(\frac{x+y}{2x+y}\right)^2\right)}{x^2}
   \]
   \[
   4 = \frac{(2x+y)^2 - (x+y)^2}{x^2}
   \]
   \[
   4 = \frac{4x^2 + 4xy + y^2 - x^2 - 2xy - y^2}{x^2}
   \]
   \[
   4 = \frac{3x^2 + 2xy}{x^2}
   \]
   \[
   4 = 3 + \frac{2y}{x}
   \]
   \[
   1 = \frac{2y}{x}
   \]
   \[
   y = \frac{x}{2}
   \]

5. Now, we need to find the area of trapezoid \(BGFC\). The area of \(BGFC\) is the difference between the area of \(\triangle ACF\) and the area of \(\triangle ABG\):
   \[
   [BGFC] = [ACF] - [ABG]
   \]
   Using the similarity ratios, we have:
   \[
   [ACF] = \left(\frac{x+y}{2x+y}\right)^2 \cdot [ADE]
   \]
   \[
   [BGFC] = \left(\frac{x+y}{2x+y}\right)^2 \cdot [ADE] - \left(\frac{x}{2x+y}\right)^2 \cdot [ADE]
   \]
   \[
   [BGFC] = \left(\left(\frac{x+y}{2x+y}\right)^2 - \left(\frac{x}{2x+y}\right)^2\right) \cdot [ADE]
   \]
   \[
   [BGFC] = \left(\frac{(x+y)^2 - x^2}{(2x+y)^2}\right) \cdot [ADE]
   \]
   \[
   [BGFC] = \left(\frac{x^2 + 2xy + y^2 - x^2}{(2x+y)^2}\right) \cdot [ADE]
   \]
   \[
   [BGFC] = \left(\frac{2xy + y^2}{(2x+y)^2}\right) \cdot [ADE]
   \]
   Substituting \(y = \frac{x}{2}\), we get:
   \[
   [BGFC] = \left(\frac{2x \cdot \frac{x}{2} + \left(\frac{x}{2}\right)^2}{(2x + \frac{x}{2})^2}\right) \cdot [ADE]
   \]
   \[
   [BGFC] = \left(\frac{x^2 + \frac{x^2}{4}}{\left(\frac{5x}{2}\right)^2}\right) \cdot [ADE]
   \]
   \[
   [BGFC] = \left(\frac{\frac{5x^2}{4}}{\frac{25x^2}{4}}\right) \cdot [ADE]
   \]
   \[
   [BGFC] = \left(\frac{5}{25}\right) \cdot [ADE]
   \]
   \[
   [BGFC] = \frac{1}{5} \cdot [ADE]
   \]
   Since \([ADE] = 36 \cdot \frac{(2x+y)^2}{x^2}\), we have:
   \[
   [BGFC] = \frac{1}{5} \cdot 180 = 45
   \]

The final answer is \(\boxed{45}\)