Help with trig identities to solve an AIME geometry question Quadrilateral $ABCD$ has side lengths $AB = 20$, $BC = 15$, $CD = 7$, and $AD = 24$, with diagonal length $AC = 25$. If we write $\angle ACB = \alpha$ and $\angle ABD = \beta$, then $\tan (\alpha + \beta)$ can be expressed in the form $-m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
What would be the general steps to solving this question?

A meta observation.   The comments discussion under the question shows how to solve this by recognizing right triangles in the picture, so that tangents of the two angles are known (and from that, the tangent of the sum of those angles).
The meta point is that had the lengths been different, we would only know $\cos \alpha$ and $\cos \beta$ to be rational numbers, and $\tan (\alpha + \beta)$ can be computed from those cosines, but there is no reason for it to be rational in that case.    
So the problem is really about the trick of noticing the special Pythagorean triangles.
