Proving that quadrilaterals are specific shapes can be shortened somewhat (as in, you don't have to literally find every vector and length and show that they're equal, or something like that).
If you have a quadrilateral ABCD then proving that

(or equivalent) proves that the shape is a parallelogram, and then showing that

(or equivalent) will prove it is a rhombus, since a rhombus is a parallelogram.
Then if you wanted to prove, say, ABCD is a square, just prove

(or equivalent) since a square is a rhombus which is a parallelogram.
Or for a rectangle, you wouldn't have to prove that that adjacent sides are equal, so you could just do

(as a rectangle is a parallelogram) and

.
If you consider only parallelograms, rhombuses, squares, and rectangles (which is pretty much all you'd need to know from memory, I doubt you'd need to recall the definition of a kite) then you can see that you'll only need at most three steps to prove a quadrilateral is one of those shapes:
For a parallelogram, you need 1 step (opposite vectors are equal)
For a rhombus, you need 2 steps (prove it is a parallelogram, and then that adjacent sides are equal)
For a rectangle, you need 2 steps (prove it is a parallelogram, and then that adjacent sides are perpendicular)
For a square, you need 3 steps (prove it is a parallelogram, that adjacent sides are equal, and that adjacent sides are perpendicular)
The order doesn't necessarily matter but you should probably start by proving a parallelogram.
A trapezium is a little different, just prove that any two sides are parallel (e.g.

for some real k)
For a cyclic quadrilateral you need opposite angles to add up to 180 degrees, so you may not always need to use vectors, but if you do have to use vectors then you can use the fact that if two angles are supplementary then their cosines add to 0 (i.e.
+cos(\theta)=0)
) and use the dot products of appropriate vectors to find the relevant cosines. And you only have to do this with one pair of angles, since it implies that the other pair of angles are also supplementary. (Note: this is assuming you already know the quadrilateral is planar. To prove a quadrilateral is planar you need to prove that any three of the vectors along its edges are linearly dependent, but that probably won't be in an exam.)
I think as far as spesh goes, that's all you need to know about proofs involving quadrilaterals.