Write the intersection condition on the line $y=1$

Since the two parabolas intersect on the line $y=1$, their common points satisfy $y=1$.

Substituting $y=1$ in both equations: $a x^2+2bx+c=0 \,\,\text{and}\,\, d x^2+2ex+f=0$

These two quadratic equations have the same roots (the $x$-coordinates of the intersection points on $y=1$).

Use the fact that quadratics with same roots are proportional

Therefore, there exists a nonzero constant $\lambda$ such that $d x^2+2ex+f = \lambda (a x^2+2bx+c)$

Comparing coefficients: $d=\lambda a, \quad e=\lambda b, \quad f=\lambda c$

Hence, $\frac da = \frac eb = \frac fc = \lambda$

Check the options involving ratios

Since $\frac da = \frac eb = \frac fc,$ the three numbers are equal.

Any three equal numbers are in:

• A.P. (common difference $0$)
• G.P. (common ratio $1$)

So both A and B are true.

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

Given $a,b,c$ are in G.P., $b^2 = ac$

Since $d=\lambda a, \quad e=\lambda b, \quad f=\lambda c,$ we get $e^2 = (\lambda b)^2 = \lambda^2 b^2 = \lambda^2 ac = (\lambda a)(\lambda c)=df$

Thus $d,e,f$ are also in G.P.

So D is true.

Check whether $d,e,f$ must be in A.P.

For A.P., we need $2e=d+f$ i.e. $2\lambda b = \lambda a + \lambda c \Rightarrow 2b=a+c$

But $a,b,c$ are given to be in G.P., not necessarily in A.P. Hence this is not always true.

So C is false.

Conclusion

The correct statements are: $\boxed{A,\;B,\;D}$

Therefore the stored answer $A$ alone is incomplete/incorrect.