We show that for a right triangle, a² + b²  =  c². Consider a right triangle

Note angle(CBD) and angle(BCD) are complementary. We have three similar triangles:

Triangle(BDC) is similar to Triangle(CDA) is similar to Triangle(BCA). Note we go from angle z to the 90 degree angle to angle 90-z.

By the properties of similar triangles, we have

    BD/CD = DC/DA = BC/CA    and   BD/BC = DC/CA = BC/BA    and    CD/BC = DA/CA = CA/BA,

where the first follows from Triangle(BDC) is similar to Triangle(CDA), the second from Triangle(BDC) is similar to Triangle(BCA), the
third from Triangle(CDA) is similar to Triangle(BCA).

Let CD = h. Substituting yields

(1)    x / h  =  h / (c-x)  =  a / b          (2)   x / a  =  h / b  =  a / c              (3)   h / a  =  (c-x) / b  =  b / c

From (2): x / a = a / c, which gives a²  =  cx.

From (3): (c-x) / b  =  b / c,  which gives b²  =  c(c-x)  =  c² - cx.

Thus, if we add these two, we find a² + b²  =  cx + c²  - cx  = c², which is what we wanted to prove.

Note we never needed to use h = CD, and in fact it was sufficient to work with Triangle(BDC) is similar to Triangle(BCA),
Triangle(CDA) is similar to Triangle(BCA).