Is this theorem valid only for parallelograms? 
"Every triangle that shares a side with a parallelogram and has its
  third vertex on the opposite side of said parallelogram's shared side,
  will have as area, the half of the parallelogram's area.

But I've seen it as a solution to a problem with a trapeze, which is not a parallelogram according to the same reference that this property comes from, so I want to know if it applies to them too? or if there is some more general form of it.
I'm translating it from Spanish so if it's not clear then I can explain it in other terms, and here's the image that comes with the explanation:

Also, this is presented as a theorem to me, or as a property, I'm using high-school geometry so I don't think they're pretty strict with the naming of them, I don't know if I know the difference anyways.
 A: This theorem only works for objects where the area is the base times the height. Since a triangle area is half the base times height its area will always be half the area. Though parallelograms are the most conventional shape with this property you could also many create wacky shapes whose area is the base times height. For example start with a rectangle (or parallelogram) and cut out any random shape on one of the horizontal sides and paste it on to the other. You now have a wacky shape whose area is base times height.
A: 
It's quite obvious that the triangle is half the parallelogram since $EF$ is parallel to the slant sides, the diagonals $EA$ and $EB$ divide the smaller parallelograms in congruent parts
Hope this helps
A: It's shouldn't work for Trapezoids; just draw one and compare the triangle against the space left. They have the same height, but different bases. This would only work if the top and bottom had the same length, which is essentially a parallelogram.
A: Yes, this holds only for parallelograms. If a is the length of the baseline and h is the height of the parallelogram, the area of the parallelogram is $a\cdot h$. The area of the triangle is $\frac{a\cdot h}2$ because the height of the triangle is the same as the height of the parallelogram.
However in a trapeze the area is $\frac{(a+c)\cdot h}2$, where c is the length of the side opposite of a. So you see that the theorem only holds if $a=c$, which is exactly when we have parallelogram.
A: It is true for the parallelogram only. Here is why: draw a line $EF$ to the side $AB$ parallel to $AD$. Note the blank and shaded triangles are equal. If the point $D$ or $C$ is moved to construct a trapezium, then the triangles no longer have equal areas.
