Prove the following relation between side lengths Let $ABC $ be any triangle , right-angled at $A$ , with $D$ any point on the side $AB$. The line $DE$ is drawn parallel to $BC$ to meet the side AC at the point E. F is the foot of the perpendicular drawn from E to BC. If $AD=x_1$ , $DB=x_2$ , $BF=x_3$, $EF=x_4$ and $AE=x_5$.  Then show that

$$\frac{x_1}{x_5} + \frac{x_2}{x_5} = \frac {x_1x_3  + x_4x_5}{x_3x_5 - x_1x_4}$$

My attempt was finding similar triangles but there are a lot of them and I really can't seem to make any progress
Note that $\tan \angle AEB=\frac{x_1+x_2}{x_5}$
and $\angle AEB = \angle AED +\color{blue}{\angle BED}=\angle AED +\color{blue}{\angle EBF}$ given that lines $DE$ and $BF$ are parallel.

Using the sum of angles formula for the tangent function we have
$$\tan \angle AEB=\frac{\tan\angle AED +\tan\angle EBF}{1-\tan\angle AED\tan\angle EBF}$$
where $\tan \angle AED=\frac{x_1}{x_5}$ and $\tan \angle EBF=\frac{x_4}{x_3}$.

Substituting results in 
$$\frac{x_1+x_2}{x_5}=\frac{\frac{x_1}{x_5}+\frac{x_4}{x_3}}{1-\frac{x_1}{x_5}\frac{x_4}{x_3}}=\frac{x_1x_3+x_4x_5}{x_3x_5-x_1x_4}$$
I think it's easy to do this using coordinates, as there are many right angles.

So, fix triangle ABC in such a way that A = origin, C = (1, 0), B = (0, b) with b > 0. Let D = (0, d) for some 0 &lt;= d &lt;= b.

Then you can calculate the coordinates of points E and F to get that 
E = (d/b, 0), F = ( (b^3 + d)/(b(b^2 + 1)), (b - d) / (b^2 + 1) ).

And then calculate all the lengths to get the required result.

It involves a bit of nasty algebra, so perhaps it's not a shortest solution, but at least a straightforward way to do without thinking too much, just performing mechanical calculations.