Below the parallelogram is obtained from square by stretching the top side while fixing the bottom.
Since area of parallelogram is base times height, both square and parallelogram have the same area.
This is true no matter how far I stretch the top side.
In below figure it is easy to see why both areas are same.
But it's not that obvious in first two figures. Any help seeing why the area doesn't change in first figure?
On
It follows from Cavalieri's Principle, or else if you know that shear transforms have determinant $1$, and hence don't change area, that's another way to see it.
I don't understand why it isn't "easy to see" that all three figures have equal bases and heights... Maybe you should just take a look at computing the difference between the images of the endpoints of the top of the squares to convince yourself.
The general form of such a transformation is
$\begin{bmatrix}1&a\\0&1\end{bmatrix}$ multiplying on the left of column vectors.)
Then you always have
$$[0,0]^T\mapsto [0,0]^T$$ $$[1,0]^T\mapsto [1,0]^T$$ $$[0,1]^T\mapsto [a,1]^T$$ $$[1,1]^T\mapsto [1+a,1]^T$$
From the first two you can see the length of the bottom horizontal line is $1$, and from the second two you can see the length of the top horizontal line is $1$. Obviously they also show the height was unchanged (since the $y$ coordinates were all preserved.)
On
In your first two figures, note that $$\text{area}(EBGH)=\text{area}(EBCH)+\text{area}(HCG)$$ and $$\text{area}(EBGH)=\text{area}(EFGH)+\text{area}(BEF).$$ But the triangles $HCG$ and $BEF$ are congruent, so have the same area. Subtracting that gives $$\text{area}(EBCH)=\text{area}(EFGH).$$ Come to think about it, this works just as well in the third figure.
On
Slice each figure by infinitely many infinitely thin horizontal layers. The area of each slice is the same as that of the corresponding slice in the original square: corresponding slices both have the same width and height and the ends can be neglected (in the given limit).
On
You can do the trick you used in your third example, where you can "move a triangle" to get to the other parallellogram multiple times.
For example:
We can do this in 2 simple steps:
step 1: just moving a triangle
step 2: moving a triangle again
On
Area of a square is base * height. Area of a right triangle is 1/2 base * height.
In first figure, base = 1, height = 1, so 1 * 1 = 1.
When pushed over, one can picture 4 triangles, 2 each above and below the unlabelled horizontal 0.5 line; each are base of 1 * height of 0.5, with area of 1/2 * (1 * 0.5), or 0.25. 4 * 0.25 = 1.
The other parallelograms are just extensions of the same.
Behold, $\phantom{proof without words}$