Given the roots of the equation are equation (a² + b²)x² – 2 (ac + bd)x + (c² + d²) = 0 are equal.

Comparing with standard quadratic equation ax² + bx + c = 0

a = (a² + b²) b = – 2(ac + bd) c = (c² + d²)

For real and distinct roots: D = 0

Discriminant D = b² – 4ac = 0

[ – 2(ac + bd)]² – 4(a² + b²)(c² + d²) = 0

4(a²c² + b²d² + 2abcd) – 4(a²c² + a²d² + b²c² + b²d²) = 0

using (a + b)² = a² + 2ab + b²

4(a²c² + b²d² + 2abcd – a²c² – a²d² – b²c² – b²d²) = 0

2abcd – a²d² – b²c² = 0

– (2abcd + a²d² + b²c²) = 0

(ad – bc)² = 0

ad = bc

Hence proved.