In $\Delta ABC, AC > AB.$ The internal angle bisector of $\angle A$ meets $BC$ at $D,$ and $E$ is the foot of the perpendicular from $B$ onto $AD$. 
In $\Delta ABC, AC > AB.$ The internal angle bisector of $\angle A$ meets $BC$ at $D,$ and $E$ is the foot of the perpendicular from $B$ onto $AD$. Suppose $AB=5,BE=4$ and $AE=3$ . Find $\Big(\frac{AC + AB}{AC - AB}\Big)ED$ .

What I Tried: Here is a picture :-

The picture only shows it. I have no other idea to try it, first that angle-chasing has no use here. Second is that I only have the required information of one right triangle, so no Pythagoras Theorem here.
I got no similar triangles, or even if I got one pair, I cannot show they are similar and so on, and in the end I am weak at Trigonometry.
Is there any way to solve this using basic techniques in Geometry? (Like angle-chasing , similarity , area , pythagorean theorem and so on) . Thank You.
Edit:- After @Michal Adamaszek's hint, I was able to do a bit of progress. (everything is there in the picture). Some congruent and similar triangles came up, but now I don't know what to do next. I need to find the value of $AC$ and $DE$ , is there any way to do it?
 A: This proof can probably be simplified or made more natural, but here is my take:

Construct $CG$ parallel to $FB$ as shown above. We have:
$$\triangle AFE \cong \triangle ABE\ (ASA), \quad \triangle AFE \sim \triangle ACG, \triangle BED \sim \triangle CGD\ (AAA)$$
So we have the following ratios:
$$AF = AB, \ \frac {AE}{EG} = \frac {AF}{FC},\ \frac {ED}{DG} = \frac {DB}{DC} = \frac {AB}{AC}$$
The last ratio is by Angle Bisector Theorem.
Hence:
$$\frac {AC+AB}{AC-AB} = \frac {AC}{CF} + \frac {AB}{CF} = \frac {AB}{CF}\left(1+\frac {AC}{AB}\right)= \frac{AF}{CF}\left(1+\frac {DG}{ED}\right)=\frac{AE}{EG}\frac {EG}{ED}=\frac{AE}{ED}$$
Giving the result:
$$\left(\frac{AC+AB}{AC-AB}\right)ED = AE$$
A: $\frac{AB}{AC}=\frac{DB}{DC}$ (This is a well-known property of the angle bisector, which certainly has some non-trigonometric proof)
$\frac{ED}{CM}=\frac{BD}{BC}$ (similarity)
$CM=\frac{3}{5}CF=\frac{3}{5}(AC-AF)=\frac{3}{5}(AC-AB)$
From there you can put it together.
A: I use a bit of trigonometry.(I would love to see something without this)
let $BC=a,CA=b,AB=c$.
By Napiers Analogy:$$\tan {\frac{B-C}{2}}=\frac{b-c}{b+c}\cot (A/2) \tag 1$$
Notice that $$\sin (A/2)=\frac{4}{5}$$
$$\cot (A/2)=? \tag 2$$
Also $\angle EDB=90+\frac{C-B}{2}$
$$\tan{\angle EDB}=\tan (90+\frac{C-B}{2})=\cot {\frac{B-C}{2}}=\frac{4}{ED} \tag 3$$
By $(1),(2),(3)$ you should be able to finish.
A: If I extend the line $BE$ to meet $AC$ at $M$, $\triangle AEB \sim \triangle AEM$ (A-A-S).
So $AM = AB = 5, EM = EB = 4$
$CM = AC - AB$
In $\triangle BMC$,
$\frac{sin \angle C}{8} = \frac{sin \angle DBE}{AC-AB}$ ...(i)
Also in $\triangle BDE$,
$\frac{sin (90^0 - \angle DBE)}{4} = \frac{\sin \angle DBE}{ED}$
$ED = 4 \tan \angle DBE$ ...(ii)
$\angle C = 90^0 - \angle CAE - \angle DBE$
$\sin C = \cos (\angle CAE + \angle DBE) = \frac{3}{5} \cos \angle DBE - \frac{4}{5} \sin \angle DBE \, \,$ ($\angle CAE = \angle BAE$) ...(iii)
$AC - AB = AC - 5 =  \frac{8 \sin \angle DBE}{\sin C}$
$AC + AB = AC - AB + 10 = \frac{8 \sin \angle DBE}{\sin C} + 10 = \frac{6 \cos \angle DBE}{\sin C}$
$\frac{AC + AB}{AC-AB}ED = 3$
EDIT:
Your modified diagram is probably the easiest solution.
$\frac{AC+AB}{AC-AB}ED = \frac{5x + 5 + 5}{5x} \times 3x \times \frac{4}{4x + 8} = 3$
