Continued proportion implies $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$

You know that $b=ka$, $c=kb$ etc so the lhs can be rewritten

$$(a^2+b^2+c^2)(k^2a^2+k^2b^2+k^2c^2) = k^2(a^2+b^2+c^2)^2$$

and the rhs can be written

$$(ka^2 + kb^2 + kc^2)^2 = k^2(a^2+b^2+c^2)^2$$

and you're done. The same trick works for the second example.

In fact the converse is also true.

This is the equality case of the Cauchy Schwarz inequality under the Euclidean Norm.

Take $\mathrm{x} = (a,b,c)$ and $\mathrm{y} = (b,c,d)$.

Geometrically, the ratio ($\sqrt{\frac{RHS}{LHS}}$) gives the cosine of the angle between $\mathrm{x}$ and $\mathrm{y}$ and is $1$ only when they are collinear (or linearly dependent).

It applies to higher dimensions too.