How to compute $BC$ in this triangle?

Question

Given the lengths of $AF$, $AE$, $AD$, $AC$, How can I calculate $BC$?

I tried to calculate $FD = \sqrt{AF^2 - AD^2}$,

$DE = \sqrt{AE^2 - AD^2}$,

$FE = DE + FD$,

$FC = AC - AF$.

But I don't know what I can do with all these lengths to compute $BC$. Any hints are appreciated.

Answer 1

Let $\alpha= \angle_{FAD}$, $\beta= \angle_{EAD}$. Then you are given $\cos\alpha= AD/AF$, and similarly $\cos\beta$. You can now compute $\cos (\alpha+\beta)$. As $\alpha+\beta = \angle_{CAB}$, you have the hypothenuse $AB$ of the right triangle $ABC$. You can finish the calculation now.

Answer 2

Trigonometry is the way to go.

We know everything about triangles $ADF$ and $ADE$. That means we know $\angle FAD$ and $\angle DAE$. From that, we know their sum, $\angle FAE=\angle CAB$. That's enough to figure out everything we need about the big triangle.