2 Right triangles, which ratio is equal to 1? Say you have a right triangle, you know the length of the 2 sides of the 90 degree corner (so you know everything, the hypotenuse and all 3 angles). Inside this triangle, you draw a line (not the height) so you create 2 new (non-similar) triangles: 1 new right triangle and another one.



Is there something that the original big right triangle (ABD) and the new smaller triangle (ABC) have in common? I am looking for a ratio that stays constant, using some property of both triangles: angles, surface, circumference, inside circle ratio, height,... I.e. the ratio of some function of alpha / (AC/AE) = that function of beta / (AD/AF), something like that, or BC/BD= ...* some function (alpha/Beta), or ... I've looked at http://en.wikipedia.org/wiki/Right_triangle, but it's not clear to me. Thanks for the help!
There is nothing that the two triangles have in common. 

Execpt one line and the right angle, but nothing else. It's a complete new triangle. There aren't any ratios or something else, that stays constant. 

That wouldn't be logical.

You know, that (using Pythagorean theorem)
$$\sqrt{AC^2-BC^2} = \sqrt{AD^2-BD^2}$$
In the $\triangle ABC$, $BE^2=AE*EC$

we, have $BE^2=BC^2-EC^2=AE*EC$

i.e. $BC^2=AC*EC$ -------(1)

In the $\triangle ACD$, $CF^2=AF*FD$

we, have $CF^2=CD^2-FD^2$

i.e. $CD^2=AD*FD$ -------(2)

Let, $a$=AREA of $\triangle ABC=\frac{1}{2}(BC)*(AB)$

Let, $b$=AREA of $\triangle ABD=\frac{1}{2}(BD)*(AB)$

we, now have, $\frac{a}{b}=\frac{BC}{BD}$ -------(3)

we, also have, $\frac{b-a}{a}=\frac{CD}{BC}$ -------(4)

$\frac{\tan\alpha}{\tan\beta}=\frac{a}{b}$

$\tan(\beta-\alpha)=\frac{1-\frac{\tan\alpha}{\tan\beta}}{\frac{1}{\tan\beta}+{\tan\alpha}}$

$\tan(\beta-\alpha)=\frac{1-\frac{a}{b}}{\frac{a}{b\tan\alpha}+{\tan\alpha}}$

$\tan(\beta-\alpha)=\frac{b-a}{\frac{a}{\tan\alpha}+{b\tan\alpha}}$

$\tan(\beta-\alpha)=\frac{(b-a)\tan\alpha}{a+b({\tan\alpha})^2}$

$\frac{\tan(\beta-\alpha)}{b-a}=\frac{\tan\alpha}{a+b({\tan\alpha})^2}$