If $DE+EF=10$ and $(AC)(EF)=40$, find BC.
I called $DE=x$, so $EF=10-x$, and we can find that $BD=\frac{20(5-x)}{x}$, but I couldn't do anymore in this question. I tried to find similar or equivalents triangles, but I didn't find.
Can someone give me a hint? Thanks for atention.
Line $FE$ is tangent to the circle as suggested in the original drawing. And the solution is not that hard.
We know the following:
$$EF=x,\ \ DE=10-x,\ \ AC=\frac{40}x\tag{1}$$
It's a well known (and easily provable) fact that:
$$EF'^2=EF^2=ED\cdot EB$$
It follows that:
$$EB=\frac{x^2}{10-x}$$
$$\tan\alpha=\frac{EB}{EF}=\frac{x}{10-x}\tag{2}$$
It's also a known fact that the angle $\alpha$ between tangent $FE$ and chord $FB$ is equal to the inscribed angle $\angle FAB$. So we know that:
$$\angle CAB=\alpha, \ \ \angle ABC=135^\circ-\alpha\tag{3}$$
Now apply law of sines to triangle $ABC$:
$$\frac{BC}{\sin\angle CAB}=\frac{AC}{\sin\angle ABC}$$
$$BC=AC\frac{\sin\angle ABC}{\sin\angle CAB}\tag{4}$$
Now replace (1) and (3) into (4):
$$BC=\frac{40}x\frac{\sin\alpha}{\sin(135^\circ-\alpha)}=\frac{40}x\frac{\sin\alpha}{\frac{\sqrt{2}}{2}\cos\alpha+\frac{\sqrt{2}}{2}\sin\alpha}$$
$$BC=\frac{40\sqrt2}{x}\frac{\sin\alpha}{\sin\alpha+\cos\alpha}=\frac{40\sqrt2}{x}\frac{\tan\alpha}{\tan\alpha+1}\tag{5}$$
Final touch: replace {2} into {5}:
$$BC=\frac{40\sqrt2}{x}\frac{\frac{x}{10-x}}{\frac{x}{10-x}+1}=4\sqrt{2}$$
Nice problem, deserved more attention.