**Proof:**

Let's consider the given equation:

(a^2 + b^2)x^2 - 2x(ac + bd) + (c^2 + d^2) = 0

We are given that the roots of this equation are equal. Let's assume this common root as 'k'.

**Roots of a Quadratic Equation:**

For any quadratic equation of the form ax^2 + bx + c = 0, the roots can be found using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our given equation, the quadratic coefficients are:

a = (a^2 + b^2)
b = -2(ac + bd)
c = (c^2 + d^2)

**Using the Quadratic Formula:**

1. Using the quadratic formula, we can find the roots of our given equation:

x = (-(-2(ac + bd)) ± √((-2(ac + bd))^2 - 4(a^2 + b^2)(c^2 + d^2))) / (2(a^2 + b^2))

Simplifying further:

x = (2(ac + bd) ± √((2(ac + bd))^2 - 4(a^2 + b^2)(c^2 + d^2))) / (2(a^2 + b^2))

2. Since we are given that the roots are equal, we can equate the discriminant (the expression inside the square root) to zero:

(2(ac + bd))^2 - 4(a^2 + b^2)(c^2 + d^2) = 0

4(ac + bd)^2 = 4(a^2 + b^2)(c^2 + d^2)

(ac + bd)^2 = (a^2 + b^2)(c^2 + d^2)

**Expanding the Equation:**

1. Expanding both sides of the equation:

a^2c^2 + 2abcd + b^2d^2 = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2

2abcd = a^2d^2 + b^2c^2 + a^2c^2

**Rearranging Terms:**

1. Rearranging the equation:

abcd = a^2d^2 + b^2c^2 + a^2c^2

2. Factoring out a common factor of 'ad' and 'bc':

ad(bc - d^2) = bc(b^2 + c^2 - ad)

**Dividing Both Sides:**

1. Dividing both sides by the common factor (bc - d^2):

ad = bc(b^2 + c^2 - ad) / (bc - d^2)

2. Simplifying the equation:

ad(bc - d^2) = bc(b^2 + c^2 - ad)

adbc - ad^3 = b^3c + bc^3 - abcd

adbc - abcd = b^3c + bc^3 - ad^3

ad(bc - bd) = bc(b^2 + c^2 - ad)

ad = bc(b^2 + c^2 - ad)