Proving that $c$ is the mean proportional between $a$ and $b$.

Question

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 $(a+c)^2:(b+c)^2=a:b$?

Answer 1

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 $a \ne b\,$.