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!
 A: 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)
$$\sqrt{AC^2-BC^2} = \sqrt{AD^2-BD^2}$$ 
A: 
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)
$\frac{\tan\alpha}{\tan\beta}=\frac{a}{b}$
$\tan(\beta-\alpha)=\frac{1-\frac{\tan\alpha}{\tan\beta}}{\frac{1}{\tan\beta}+{\tan\alpha}}$
$\tan(\beta-\alpha)=\frac{1-\frac{a}{b}}{\frac{a}{b\tan\alpha}+{\tan\alpha}}$
$\tan(\beta-\alpha)=\frac{b-a}{\frac{a}{\tan\alpha}+{b\tan\alpha}}$
$\tan(\beta-\alpha)=\frac{(b-a)\tan\alpha}{a+b({\tan\alpha})^2}$
$\frac{\tan(\beta-\alpha)}{b-a}=\frac{\tan\alpha}{a+b({\tan\alpha})^2}$
