# Finding the Perimiter of a right Triangle given an interior angle bisector and exterior angle bisector

I've been working on this problem for my Foundations of Geometry class since yesterday and I still can't figure it out. Any help you could give would be appreciated. Here's the problem:

ABC is a right triangle with its right angle at C. The bisector of angle B intersects ray AC at D, and the bisector of the exterior angle at B intersects line AC at E. If BD = 15 and BE = 20, what is the Perimeter of Triangle ABC?

Hint: Draw a picture. It is easy to see that $\triangle BDE$ is right-angled at $E$. So now we know the hypotenuse $DE$.
But $\triangle BDC$ is similar to $\triangle BDE$. Since we know the side $BD$ of $\triangle BDC$, and we know everything about $\triangle BDE$, we know everything about $\triangle BDC$.
Now there are various ways we can branch. One way is to go trigonometric (with exact values). We know everything about $\triangle BDE$, so we know the trig. functions for $\angle DBC$. But $\angle ABC$ is twice $\angle DBC$, so now we know, via the double angle formula, the trig. functions of $\angle ABC$.
But now since we know side $BC$, we know the other sides too.
Remark: A fancier fact to use, if we want to be more Euclidean, is that the ratio of $BC$ to $CD$ is the same as the ratio of $BA$ to $DA$. This theorem, which is in Euclid, nowadays has a name like "angle bisector" theorem. I prefer crude.