# 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!

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$C$ can be any point on $BD$ and you can draw a diagram which works. The two triangles have side in common and are both right-angled. AB/AB=1. It would help to have some motivation for the question - why do you think there might be something the same? – Mark Bennet Aug 23 '12 at 14:16
If you throw in $\alpha$ and $\beta$, there is a large number of relationships. One can produce a few without mentioning angles. There seems to be no point in producing a random list without knowing what the aim is. – André Nicolas Aug 23 '12 at 17:49
Hi André, thanks for responding. The aim is to use this in a filter, and I need to be able to treat a variety of movements of a variable in the same way: it doesn't matter to this filter how high the variable goes up, so I must calculate something from a reference point that allows me to classify all these moves in the same category so I can ignore the difference in vertical movement. For this, I need to compare the previous movement with the current movement, and if the "ratio" that I want to calculate stays the same, then I don't need to adjust anything. – MisterH Aug 23 '12 at 19:48

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) $$\sqrt{AC^2-BC^2} = \sqrt{AD^2-BD^2}$$

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There is one candidate for $C$ which would make the two triangles similar, provided $AB<BD$ – Arthur Aug 23 '12 at 13:55
Shouldn't there be some sort of relationship between (alpha/beta)*AB and BC/BD? – MisterH Aug 23 '12 at 13:57
If you would think of it from a Cartesian coordinate point of view, if you start at the origin and you have a certain angle alpha, and you draw a line of a certain length, what would the angle beta be to cover the same horizontal distance (AB in my example)? – MisterH Aug 23 '12 at 14:21

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)

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

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

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

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

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

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

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