Finding an alternate solution to a plane-geometry problem. I'm having some trouble finding an alternate solution to the following problem:

Let us have a right-angled triangle $ABC$ ($A$ being the right angle vertex). We draw an altitude to the hypotenuse creating point $D$ and an angle bisector of the vertex angle C which intersected with $AB$ gives us another point $E$. Prove that $|AE| \cdot|AC|=|BE| \cdot |CD|. $

I do know how to solve the problem with the use of analytic geometry, however I'm curious wether there's some less computationally demanding way. Any help is appreciated.

 A: Let $ \tan \frac{C}{2}  = t$.  Let $AC = 1 $.
Show that

*

*$AE = t$

*$BE = \tan C - t = \frac{ 2t } { 1-t^2 }  - t $

*$CD = \cos C = \frac{1-t^2}{1+t^2}$
Hence, $ |AE | \times |AC | = |BE| \times |CD |$
A: Here is a straightforward proof using two theorems from Euclidean geometry.
Theorem 1:  If an angle of a triangle is bisected by a straight line cutting the base, then the segments of the base have the same ratio as the remaining sides of the triangle.  (You can find this in Euclid's Elements, Book VI, Proposition 3.) In this case, we conclude that $$\frac{|AE|}{|BE|} = \frac{|AC|}{|BC|}$$
or, equivalently,
$$|AE| \cdot |BC| = |BE| \cdot |AC|$$
Theorem 2: The altitude drawn to the hypotenuse of a right triangle divides the triangle into two similar triangles.  In this case, we conclude that $\triangle ADC \sim \triangle BDA$, and hence
$$\frac{|AC|}{|BC|} = \frac{|CD|}{|AC|} $$
Multiply these two equations together and you get the result you want.
A: $\frac{AE}{BE}=\frac{AC}{BC}$ by the angle bisector theorem, which is true in any triangle. Then $\frac{AC}{BC}=\cos C=\frac{CD}{AC}$.
