1. Use the condition that $a,b,c$ are in G.P.

Since $a,b,c$ are in G.P., we have $b^2=ac.$ Because the numbers are distinct, the quadratic $ax^2+2bx+c=0$ is not a perfect square with equal coefficients, but the G.P. relation is the key.

2. Find the common root of the first quadratic.

For $ax^2+2bx+c=0,$ the discriminant is $\Delta=(2b)^2-4ac=4(b^2-ac)=0.$ So the quadratic has a repeated root: $x=-\frac{2b}{2a}=-\frac{b}{a}.$

Thus the common root of both equations is $x=-\frac{b}{a}.$

3. Apply the common root condition to the second quadratic.

Since $x=-\frac{b}{a}$ is also a root of $dx^2+2ex+f=0,$ substitute it: $d\left(-\frac{b}{a}\right)^2+2e\left(-\frac{b}{a}\right)+f=0.$

So, $d\frac{b^2}{a^2}-2e\frac{b}{a}+f=0.$

Multiply by $a^2$: $db^2-2abe+a^2f=0.$

Using $b^2=ac$, $acd-2abe+a^2f=0.$

Now divide by $a$: $cd-2be+af=0.$

Rearrange: $af+cd=2be.$

4. Interpret this relation.

Divide by $ac$: $\frac{af}{ac}+\frac{cd}{ac}=2\frac{be}{ac}.$

That is, $\frac{f}{c}+\frac{d}{a}=2\frac{e}{b},$ using $ac=b^2$ in the middle term: $2\frac{be}{ac}=2\frac{e}{b}.$

Hence, $\frac{d}{a}+\frac{f}{c}=2\frac{e}{b}.$

This is exactly the condition that $\frac{d}{a},\;\frac{e}{b},\;\frac{f}{c}$ are in A.P.

5. Check the options.

• A: $\dfrac da,\dfrac eb,\dfrac fc$ are in G.P. — not implied.
• B: $d,e,f$ are in A.P. — not implied.
• C: $d,e,f$ are in G.P. — not implied.
• D: $\dfrac da,\dfrac eb,\dfrac fc$ are in A.P. — correct.

Therefore, the correct option is $\boxed{D}.$