A proof of Pythagoras Theorem

Here is a potentially new proof of the Pythagoras theorem I discovered recently. I could not find any prior proofs like these (including in the cut-the-knot pythagorean proof dump), so if you have seen it somewhere, please let me know.

ABC is the right triangle with sides a,b,c.

AD is the angular bisector of CAB and DE is the perpendicular from D to AB.

CD = x, DB = a-x.

It is easily seen that triangles ACD and ADE are congruent and so AE = AC = b and BE = c-b

Area of ACD + Area of ADB = Area of ABC gives

$bx + cx = ab $

And so $x = ab/(b+c)$

Now triangles DEB and ABC are similar and so

$$DE/AC = BE/BC$$

i.e.

$$x/b = (c-b)/a$$

$$a/(b+c) =(c-b)/a$$

Which gives

$$ a^2 = c^2 - b^2$$