Proving that $c$ is the mean proportional between $a$ and $b$. 
If $a≠b$ and $a:b$ is the duplicate ratio of $(a+c):(b+c)$, prove that $c$ is the mean proportional between $a$ and $b$.

My attempt:
I have assumed that $c$ is the mean proportional between $a$ and $b$, expressed $a$ and $c$ in terms of $b$ and then substituted them in $(a+c)^2:(b+c)^2$ and $a:b$ to get equal expressions on L.H.S and R.H.S, i.e., i have worked in a reverse manner.
My question:
How to prove this starting from $(a+c)^2:(b+c)^2=a:b$?

Hint:
$$
\require{cancel}
\begin{align}
\frac{a}{b}=\frac{(a+c)^2}{(b+c)^2} \quad & \implies \quad a(b+c)^2=b(a+c)^2 \\
 & \iff \quad a(b^2+2bc+c^2)=b(a^2+2ac+c^2) \\
 & \iff \quad ab^2 +\cancel{2abc}+ac^2-a^2b-\cancel{2abc}-bc^2=0 \\
 & \iff \quad ab(b-a)+(a-b)c^2=0 \\
 & \iff \quad (a-b)(c^2-ab)=0
\end{align}
$$
All that's left is to remember that $a \ne b\,$.
