(Seemingly) simple trigonometry problem Given the diagram below, I'm trying to determine the value of $x$ in terms of $a$, $b$ and $c$.
I've already run this with a few of my colleagues. It seems obvious that the problem has only one solution, but we can't find it… is trigonometry the way to go, as we thought?

If anyone is interested, the original problem I was trying to solve was this: I was tracing a letter Z (in Inkscape) and wanted to make sure the stroke width was constant. In order to do this, I need to determine the horizontal and vertical positions for the start and end of the diagonal bar, as illustrated here:

 A: Computing total area in two ways, we have  $$ \frac{1}{b}(c+x) = \frac{1}{2}b(c-x) + a\sqrt{b^2+(c-x)^2} $$
and hence
$$\frac{bx}{a} = \sqrt{b^2+(c-x)^2} $$
The rest of proof is same as that in Parcly Taxel's solution.
A: 
The trick here is to shift the perpendicular between the two diagonal lines upwards so that two similar triangles are obtained. Then, by applying the Pythagorean theorem, we obtain
$$\frac ax=\frac b{\sqrt{b^2+(c-x)^2}}$$
which may be manipulated to obtain a quadratic polynomial for $x$:
$$a\sqrt{b^2+(c-x)^2}=bx$$
$$a^2(b^2+(c-x)^2)=b^2x^2$$
$$b^2+c^2-2cx+x^2=\frac{b^2}{a^2}x^2$$
$$\left(1-\frac{b^2}{a^2}\right)x^2-2cx+b^2+c^2=0$$
After solving, you need to check whether the obtained $x$ values are sensible – they have to lie within 0 and $c$.
A: Let's define an angle $\alpha$ which defines the line between the lines noted a and b. Then we have $x=a/\cos\alpha$ and $c=x+b\tan\alpha$. Using $x=\cos^{-1}(\alpha/x)$, $\tan\alpha=\sin\alpha/\cos\alpha$ and $\sin\cos^{-1}y=\sqrt{1-y^2}$, we get $c=x+b\sqrt{1-\alpha^2/x^2}/(a/x)$. Simplifying that we get Parcly Taxel's solution.
A: In the diagram there is a parallelogram on the left, and a right triangle on the right, and I denote by $y$ the oblique segment which is both a side of the parallelogram and the hypotenuse of the triangle.
By computing the area of the parallelogram in two ways, we see:
$$bx = ay$$
but $y$ is also easy to determine from Pythagoras applied to the right triangle, so plugging in:
$$bx = a\sqrt{b^2+(c-x)^2}$$
which is the same equation all the other answer arrive at.
