(Seemingly) simple trigonometry problem

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 $c$.

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.