If a, b, c, d are in proportion, prove that (a squared+b squared=c squared)(b squared+c squared+d squared)=(ab+bc+cd) squared?

Answer 1

Let's substitute:

a=1, b=2, c=3, d=4

(1^2+2^2)=3^2*(2^2+3^2+4^2)=(1*2+2*3+3*4)^2

5 =9(4+9+16) =(2+6+12)^2

5 =261 =400

Not with this formula.

Answer 2

a / b = c / d

Allow a / b = c / d = p

(a² +b² + c²)(b² + c² + d²)

Now a² = p²b², b² = p²c², and c² = p²d²

So

(p²b² + p²c² + p²d²)(b² + c² + d²)

p²(b² + c² + d²)²

(pb² + pc² + pd²)²

(pbb + pcc + pdd)²

((ab + bc + cd)²

QED!