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