If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between (a² + b²) and (b² + c²).

Given, b is the mean proportional between a and c then,

b² = ac.    [….Eq 1]

For (ab + bc) to be the mean proportional between (a² + b²) and (b² + c²) following condition must be satisfied,

(ab + bc)² = (a² + b²)(b² + c²)

Solving L.H.S. first,

$\Rightarrow (ab + bc)^2 \\[0.5em] = a^2b^2 + 2ab^2c + b^2c^2 \\[0.5em]$

Putting value of b² from equation 1:

$= a^2(ac) + 2ac(ac) + c^2(ac) \\[0.5em] = a^3c + 2a^2c^2 + ac^3 \\[0.5em] = ac(a^2 + c^2 + 2ac) \\[0.5em] = ac(a + c)^2$

Now, solving R.H.S. ,

$\Rightarrow (a^2 + b^2)(b^2 + c^2) \\[0.5em] = (a^2b^2 + a^2c^2 + b^4 + b^2c^2) \\[0.5em]$

Putting value of b² from equation 1:

$= (a^2(ac) + a^2c^2 + (ac)^2 + (ac)(c^2) \\[0.5em] = a^3c + a^2c^2 + a^2c^2 + ac^3 \\[0.5em] = a^3c + 2a^2c^2 + ac^3 \\[0.5em] = ac(a^2 + 2ac + c^2) \\[0.5em] = ac(a + c)^2$

Since, L.H.S. = R.H.S. = ac(a + c)² hence,
(ab + bc) is the mean proportional between (a² + b²) and (b² + c²).