middle school question on geometry 
As you can see from the picture, the angle $A$ is $90^\circ$, and the segments $BD$ and $CE$ (which intersect at $F$) are angle bisectors of the angles $B$ and $C$, respectively. When the length of $CF$ is $\frac72$ and and the quadrilateral $BCDE$ has an area of $14$, what is the length of $BC$?
This is supposedly a middle-school question, appreciate any help.

This is not a middle school level answer. But perhaps a complicated answer is better than no answer at all. Start by choosing coordinates as follows:
$$
A=\begin{pmatrix}0\\0\end{pmatrix}\quad
B=\begin{pmatrix}b\\0\end{pmatrix}\quad
C=\begin{pmatrix}0\\c\end{pmatrix}\quad
D=\begin{pmatrix}0\\d\end{pmatrix}\quad
E=\begin{pmatrix}e\\0\end{pmatrix}
$$
Then the area condition becomes
$$\tfrac12bc-\tfrac12de=14\tag{1}$$
For the distance condition, you have to compute
$$F=\frac{1}{bc-de}\begin{pmatrix}be(c-d)\\cd(b-e)\end{pmatrix}$$
by intersecting $BD$ with $CE$. As an alternative, you might intersect one of these lines with the bisector $x=y$, but I'll leave the above for now. From that you get
\begin{align*}
\lVert F-C\rVert &= \tfrac72 \\
\lVert F-C\rVert^2 &= \left(\tfrac72\right)^2 \\
\left(\frac{be(c-d)}{bc-de}\right)^2+\left(\frac{cd(b-e)}{bc-de}-c\right)^2
&= \left(\tfrac72\right)^2 \\
%\left(\frac{1}{bc-de}\right)^2\left((be(c-d))^2 + (cd(b-e)-c(bc-de))^2\right)
%&= \left(\tfrac72\right)^2 \\
\bigl(be(c-d)\bigr)^2 + \bigl(cd(b-e)-c(bc-de)\bigr)^2
&= \left(\tfrac72(bc-de)\right)^2 \tag{2}
\end{align*}
The most difficult part to formulate is probably the angle bisector conditions. Start at the double angle formula for the tangens:
\begin{align*}
\tan2\theta &= \frac{2\tan\theta}{1-\tan^2\theta} \\
\frac cb &= \frac{2\frac db}{1-\left(\frac db\right)^2} \\
\frac cb &= \frac{2bd}{b^2-d^2} \\
c(b^2-d^2) &= 2b^2d \tag{3} \\
b(c^2-e^2) &= 2c^2e \tag{4}
\end{align*}
Now you have four (non-linear) equations $\text{(1)}$ through $\text{(4)}$ in four variables $b$ through $e$. Eliminate variables (e.g. using resultants) to obtain the solution:
$$
b=\frac{120}{17} \quad
c=\frac{161}{34} \quad
d=\frac{840}{391} \quad
e=\frac{644}{255}
$$
Then you can compute
$$
\lVert B-C\rVert=\sqrt{b^2+c^2}=\frac{17}2
$$

