Geometry

We can use similar triangles to prove this identity.

Let's draw a diagram of the right triangle ABC with the altitude CH:

```
       C
      /|
     / |
  H /  | 
   /   | 
  /____|
 A  B
```

Since CH is perpendicular to AB, we have:

angle HCB = 90 - angle C = angle A
angle HBC = 90 - angle C = angle B

Therefore, triangles CHB and CAB are similar. That is:

CH / CA = CB / CB

or equivalently:

CH / a = b / (a + b)

Cross-multiplying and simplifying, we get:

CH = ab / (a + b)

Now, let's consider the right triangle HCB:

```
   C
  /|
 / |
/H |
/__|
B  A
```

Since angle HCB = angle A, we have:

angle HBC = angle C

Therefore, triangles HCB and ABC are similar. That is:

HC / AB = BC / AC

or equivalently:

HC / (a + b) = b / c

Cross-multiplying and simplifying, we get:

HC = b(a + b) / c

Now, let's use the Pythagorean theorem in triangle ABC:

c^2 = a^2 + b^2

Adding 2ab to both sides, we get:

c^2 + 2ab = a^2 + 2ab + b^2

Simplifying, we get:

c^2 + 2ab = (a + b)^2

Taking the fourth power of both sides, we get:

(c^2 + 2ab)^2 = (a + b)^4

Substituting our expressions for CH and HC, we get:

(ab / (a + b) + b(a + b) / c)^2 = (a + b)^4

Simplifying, we get:

[(ab + (a + b)^2 / c)^2] / (a + b)^2 = (a + b)^2

Multiplying both sides by (a + b)^2 and simplifying, we get:

(ab + (a + b)^2 / c)^2 = (a + b)^4

Substituting x = a + b and h = ab / (a + b), we get:

(x^2 / c + h)^2 = x^4

Expanding both sides and simplifying, we get:

(x^2 / c)^2 + 2x^2h / c + h^2 = x^4

Multiplying both sides by c^2 and simplifying, we get:

x^4 - 2x^2h - c^2h^2 = 0

Using the quadratic formula, we get:

x^2 = h ± sqrt(h^2 + c^2h^2)

Since x > 0, we must have:

x^2 = h + sqrt(h^2 + c^2h^2)

Substituting back our expressions for x and h, we get:

(a + b)^2 = ab / (a + b) + sqrt(ab^2 / (a + b)^2 + c^2 ab^2 / (a + b)^2)

Multiplying both sides by (a + b)^2 and simplifying, we get:

(a + b)^4 = ab(a + b)^2 + c^2 ab^2

Therefore, we have proved that:

(x + h)^2*(y + h)^2 = (a + b)^4.