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 $AD$.

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:

$$\left(\frac{AC+AB}{AC-AB}\right)ED = AE$$

$\frac{AB}{AC}=\frac{DB}{DC}$ (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.