Hi,

Given two right angle triangles sharing a common side AC.

Call them ABC, and ADC.

First triangle angle ACB is $\displaystyle \frac{ \pi }{2}-\theta$

angle ABC is a right angle.

Second triangle angle CAD is $\displaystyle \frac{ \pi }{2}-\alpha$

angle ADC is a right angle.

Why is the following statement true.

$\displaystyle

\frac{BC}{cos(\frac{\pi}{2}-\theta)} = \frac{AD}{cos(\frac{\pi}{2}-\alpha)}

$

I don't see how this can hold when $\displaystyle \theta \neq \alpha$

I tried using pythagoras, but that doesnt' work.

Perhaps I am missing something about the angle of refraction.