# 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.

• If it's not a middle-school question, don't appreciate any help? – Paul Hurst May 30 '14 at 23:46
• ok my bad. The angle sign made my paragraph dissappear. – user59768 May 30 '14 at 23:47
• This question should be reopened now. – David H May 31 '14 at 3:29

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

$$\lVert B-C\rVert=\sqrt{b^2+c^2}=\frac{17}2$$