A cuboid ABCDEFGH has base ABCD. AE is perpendicular to the base. Given AB=16, AE=6. Find a) the projection of line BE onto plane ABCD. b) i) the angle between AD and BF. ii) the angle between BE and

a) The projection of the line BE on the plane ABCD is the line segment connecting the projection of E to the plane and the point B, which is already on the plane. Since E is vertically above A, the projection of E onto plane ABCD is A.
Therefore, the projection of line BE on the plane ABCD is BA.

The projection of line BE on the plane ABCD is $\boxed{\text{BA}}$.

b)

i) To find the angle between lines AD and BF, we can translate one of the lines so that they intersect. Line BF is parallel to line AE. So, the angle between AD and BF is the same as the angle between AD and AE.
Since AE is an edge perpendicular to the base plane ABCD, AE is perpendicular to any line in the plane ABCD that passes through A. AD is such a line.
Therefore, the angle between AD and AE is $90^\circ$.

The size of the angle between AD and BF is $\boxed{90.0^\circ}$.

ii) The angle between line BE and the plane ABCD is the angle between BE and its projection onto the plane, which is BA (from part a). This is $\angle EBA$.
Consider the right-angled triangle EAB, where $\angle EAB = 90^\circ$.
We are given:
AB = 16 cm
EA = CF = 6 cm (height of the cuboid)

We can use the tangent function:

$$\tan(\angle EBA) = \frac{\text{opposite}}{\text{adjacent}} = \frac{EA}{AB}$$ $$\tan(\angle EBA) = \frac{6}{16} = \frac{3}{8}$$ $$\angle EBA = \arctan\left(\frac{3}{8}\right)$$ $$\angle EBA \approx 20.556^\circ$$

Correct to 1 decimal place, the angle is $20.6^\circ$.

The angle between line BE and the plane ABCD is $\boxed{20.6^\circ}$.

iii) The planes HBCE and BCFG are adjacent faces of the cuboid. The line of intersection of these two planes is BC.
To find the angle between the planes, we consider two lines, one in each plane, that are perpendicular to the line of intersection (BC) at the same point. Let's choose point C on BC.
In plane BCFG, the line CF is perpendicular to BC (since BCFG is a rectangle).
In plane HBCE, the line CH is perpendicular to BC (since HBCE is a rectangle).
The angle between the planes is the angle between CF and CH.
Since it is a cuboid, the edges meeting at a vertex are mutually perpendicular. Therefore, CF is perpendicular to CH.

The angle between planes HBCE and BCFG is $\boxed{90.0^\circ}$.

c) Point N is the midpoint of EF.
The length of edge EF is equal to AB, so EF = 16 cm.
Since N is the midpoint of EF, NF = $\frac{1}{2} \times EF = \frac{1}{2} \times 16 = 8$ cm.
Consider the triangle BFN. The edge BF is perpendicular to the top face EFGH, so BF is perpendicular to NF. Thus, $\triangle BFN$ is a right-angled triangle with the right angle at F.
We are given:
BF = CF = 6 cm (height of the cuboid)
NF = 8 cm

Using the Pythagorean theorem:

$$BN^2 = BF^2 + NF^2$$ $$BN^2 = 6^2 + 8^2$$ $$BN^2 = 36 + 64$$ $$BN^2 = 100$$ $$BN = \sqrt{100}$$ $$BN = 10 \text{ cm}$$

Correct to 1 decimal place, the length BN is 10.0 cm.

The length BN is $\boxed{10.0 \text{ cm}}$.