$\because a,b,c$ are in G.P. $\Rightarrow b^2=ac$ $\dots \left(i\right)$
Equation $a{x}^2+2bx+c=0$
Hence, $D=4b^2-4ac=0$ $\left(\because b^2=ac\right)$
$\Rightarrow$ equation has equal roots $\Rightarrow$ roots are $=\frac{-b}{a},\frac{-b}{a}$ ( $\because$ sum of roots $\alpha +\alpha =\frac{-2b}{a}$ )
$\because$ Both equations have one root in common.
$\therefore$ Common root is $-\frac{b}{a}$. It will satisfy equation.
$\therefore d{\left(\frac{-b}{a}\right)}^2+2e\left(-b\right)a+f=0\Rightarrow d{b}^2-2aeb+a^{2}f=0$
$\Rightarrow d\left(ac\right)-2aeb+a^{2}f=0$ $\left(\because b^2=ac\right)$
$\Rightarrow dc-2aeb+af=0$
$\Rightarrow \frac{dc}{ac}-\frac{2eb}{ac}+\frac{af}{ac}=0$ (dividing by $ac$ )
$\Rightarrow \frac{d}{a}-\frac{2eb}{b^2}+\frac{f}{c}=-\Rightarrow \frac{d}{a}+\frac{f}{c}=\frac{2e}{b}$
$\Rightarrow \frac{d}{a},\frac{e}{b},\frac{f}{c}$ are in A.P.