If $x$, $y$ and $z$ are in continued proportion, prove that $\;$ $\dfrac{\left(x + y\right)^2}{\left(y + z\right)^2} = \dfrac{x}{z}$
$x, \; y, \; z$ are in continued proportion.
$\implies$ $\dfrac{x}{y} = \dfrac{y}{z} = k \;$ (say)
$\implies$ $y = z k$ $\;\;\; \cdots \; (1)$
and $\;$ $x = yk = zk^2$ $\;\;\; \cdots \; (2)$
$\begin{aligned}
LHS & = \dfrac{\left(x + y\right)^2}{\left(y + z\right)^2} \\\\
& = \dfrac{\left(zk^2 + zk\right)^2}{\left(zk + z\right)^2} \;\;\; \left[\text{in view of equations (1) and (2)}\right] \\\\
& = \dfrac{\left[zk \left(k + 1\right)\right]^2}{\left[z \left(k + 1\right)\right]^2} \\\\
& = k^2 \;\;\; \cdots \; (3) \\\\
RHS & = \dfrac{x}{z} \\\\
& = \dfrac{zk^2}{z} = k^2 \;\;\; \cdots \; (4)
\end{aligned}$
$\therefore \;$ From equations $(3)$ and $(4)$, $\;\;$ $LHS = RHS$
i.e. $\;$ $\dfrac{\left(x + y\right)^2}{\left(y + z\right)^2} = \dfrac{x}{z}$
Hence proved.