Geometric Proof?

Here's a geometric proof of the Pythagorean Theorem using similar triangles :

Let  ABC  be a righ triangle with a right angle at C

And CD is an altitude  with right angles ADC  and BDC

We want to prove that  AC^2  + BC^2 =  AB^2

So...looking at triangles ABC  and ADC

Angle A  = angle A

Angle ACB  = angle CDA

So...by AA  congruency.....Δ ACB ~Δ ADC  →  AC /AB  = AD/ AC  →  AC^2  = AB * AD

Similary  Δ ACB ~ ΔCDB → AB / BC = BC / DB  →  BC^2  = AB * DB

So

AC^2  + BC^2  = AB* AD   + AB * DB

AC^2 + BC^2  = AB ( AD + DB)

AC^2 + BC^2  = AB (AB)

AC^2 + BC^2  = AB^2