Show that $\frac{\log_aN-\log_bN}{\log_bN-\log_cN}=\frac{\log_aN}{\log_cN}$

Show that $$\dfrac{\log_aN-\log_bN}{\log_bN-\log_cN}=\dfrac{\log_aN}{\log_cN}$$ where $a,b$ and $c$ are positive and are consecutive terms of а geometric sequence, $a\ne1,b\ne1,c\ne1,N>0,N\ne1$.

$a,b$ and $c$ are consecutive terms of a geometric sequence if and only if $b^2=ac, b=\sqrt{ac}$. Then the LHS is $$\dfrac{\log_aN-\log_\sqrt{ac}N}{\log_\sqrt{ac}N-\log_cN}=\dfrac{\log_aN-2\log_{ac}N}{2\log_{ac}N-\log_cN}$$ This seems useless. What can we do? What is the intuition that will lead to the solution? Thank you!

$$\dfrac{\log_aN-\log_bN}{\log_bN-\log_cN}-\dfrac{\log_aN}{\log_cN}$$ $$=\dfrac{\frac{1}{\log_Na}-\frac{1}{\log_Nb}}{\frac{1}{\log_Nb}-\frac{1}{\log_Nc}}-\frac{\frac{1}{\log_Na}}{\frac{1}{\log_Nc}}$$ $$=\dfrac{\log c\log b-\log a \log c}{\log a\log c-\log a \log b}-\frac{\log c}{\log a}$$ (where all logs are base $N$) $$=\frac{\log c}{\log a}\left[\frac{\log b-\log a}{\log c - \log b}-1\right]$$

But $$\log b -\log a =\log c-\log b = \log r,$$ where $r$ is the common ratio. Therefore the expression is zero as required.