No, this is only an approximation based on the assumption that the L/D ratio is fairly high. Imagine that the glider has a gigantic drogue chute attached, so that the flight path is aimed almost straight down. Almost all the aircraft's weight will be borne by the drag vector, and almost none of the aircraft's weight will be borne by the lift vector. The weight vector will be almost parallel to the flight path and the total "power" exerted by gravity will be equal to weight * descent speed, which will be almost equal to weight * airspeed.
If the glider (w/ drogue chute still attached) now powers up a powerful motor (guess it was a motorglider) and tries to fly horizontally, why should we expect the minimum power needed to maintain altitude to be equal to weight * the descent speed from the power-off case?
Here is the key to seeing why when the L/D ratio is poor, the horizontal flight case takes much more power than the gliding case:
When the L/D ratio is high, then when we draw the L-D-W vector triangle, L is very nearly as large as W.
But when the L/D ratio is very low-- e.g. the drogue chute case-- then when we draw the L-D-W vector triangle, D is nearly as large as W, and L is very much smaller than W.
When we start from the gliding case w/ a low L/D ratio, and then add power to maintain horizontal flight, what must happen?
Take a pencil and paper and draw the L-D-W vector triangle for such a case-- where the L/D ratio is very poor. Now to change the picture to represent the powered case, you have to take the L and D vectors, keeping the 90-degree angle between then, and swing them upwards until the D vector is horizontal and the L vector is vertical. Then you must EXPAND the L and D vectors, keeping the same proportion between them, until the L vector is fully as large as the W vector, so L is equal and opposite to W. Assuming that our angle-of-attack and lift coefficient are staying constant, this EXPANSION of the L and D vectors represents a huge increase in airspeed compared to the gliding case. That's the reason that we'll need MUCH more engine power to maintain level flight, than gravity was providing in the glide. In both cases, power must be equal to airspeed times drag (which in the gliding case is also equal to weight times sink rate), but when the L/D ratio is very poor, the airspeed and the magnitude of the drag vector are both MUCH higher in the horizontal-flight case than in the gliding case.
This answer undoubtedly could be improved by the addition of correctly drawn diagrams for the case where the L/D ratio is very poor, showing both the closed L-D-W vector triangle for the unpowered case, and the L-D-W-T diagram for the powered case, preserving a constant L/D ratio in both diagrams.