Let the radius of the base of the cone(DB) = R

Height of the base of the cone(AD) = H

Slant height of the base of the cone(AB) = L

Again, let the radius of the base of the cone(GE) = r

Height of the base of the cone(AG) = h

Slant height of the base of the cone(AE) = l

Now, from the figure

Height of the remaining part of the cone GD = H - h

Given that the right circular cone is divided by a plane parallel to its base in two equal volumes,

So, the volume of the small cone is half of the volume of the whole cone.

=> πr² h/3 = (1/2)*πR² H/3

=> r² h = R² H/2

=> r² h/R² H = 1/2

=> r² /R² = H/2h ..............1

Again, we know that EF || BC,

So, ΔAEG ∼ ΔABD

=> GE/DB = AG/AD

=> r/R = h/H

=> (r/R)² = (h/H)²

=> r² /R² = h² /H² ...........2

From equation 1, we get

      H/2h = h² /H²

=> H/2h = (h/H)²

=> (1/2)*(h/H) = (h/H)²

 =>  1/2 = (h/H)³

=> h/H = (1/2)^1/3

=> h/H = 1/2^1/3

=> H = h*2^1/3

Now, h/(H - h) = h/(h*2^1/3 - h)

=> h/(H - h) = h/{h(2^1/3 - 1)}

=> h/(H - h) = 1/(2^1/3 - 1)

=> h : (H - h) = 1 : (2^1/3 - 1)

So, the ratio of the line segment into which the axis of the cone is divided by the plane is 1 : (2^1/3 - 1)