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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.

enter image description here

This is true no matter how far I stretch the top side.

enter image description here

In below figure it is easy to see why both areas are same. enter image description here

But it's not that obvious in first two figures. Any help seeing why the area doesn't change in first figure?

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9 Answers 9

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Behold, $\phantom{proof without words}$

picture

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    $\begingroup$ Yeah: that picture proof is pretty robust. Has all the elements to formulate a synthetic argument too. $\endgroup$
    – rschwieb
    Jul 25, 2019 at 2:25
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    $\begingroup$ Ahahah. I wish I could vote twice just for "Behold!" $\endgroup$
    – Tasos Papastylianou
    Jul 25, 2019 at 10:21
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    $\begingroup$ Gentlemen, BEHOLD (cit. Dr. Weird) $\endgroup$
    – TheVal
    Jul 27, 2019 at 10:43
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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).

enter image description here

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    $\begingroup$ Seems circular. You would like to say exactly the same thing about the thin slices as you would for the whole, but you don't have it yet (unless you assume the conclusion.) $\endgroup$
    – rschwieb
    Jul 24, 2019 at 19:26
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    $\begingroup$ Agree with @rschwieb . You may want to put something in about the thin horizontal slices being just straight up rectangles instead of sheared rectangles on top of one another, so that they could've also been gotten in the original square, and as we make them infinitely small, the "empty space" due to having to shift over the ones getting stacked on top approaches 0? $\endgroup$
    – joshuaronis
    Jul 24, 2019 at 20:02
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    $\begingroup$ Not circular at all, though I clarified the answer to include this. $\endgroup$
    – David G. Stork
    Jul 24, 2019 at 21:49
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    $\begingroup$ @DavidG.Stork Taken at face value it was circular... but I see now you have expanded it to a sketch of Cavalieri’s principle, yes. $\endgroup$
    – rschwieb
    Jul 25, 2019 at 1:07
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    $\begingroup$ An alternative version of this is to take the parallelogram and slice it into sufficiently many horizontal parallelogram strips so that each strip can individually be sheared into rectangles according to the OP's "easy to see why" picture. These rectangles can be slid around to form a big rectangle. $\endgroup$
    – Kyle Miller
    Jul 25, 2019 at 3:59
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Just for completeness, here is another method of disection that proves the result. parallelogram disection

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    $\begingroup$ Beautiful! Looks the distance between the lines approaches $0$ as FG goes to the right extreme.. making the width of the slices in the square also approach $0$. $\endgroup$
    – AgentS
    Jul 26, 2019 at 8:38
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    $\begingroup$ @rsadhvika The width of the slices approaches zero, but, at the same time, the number of slices approaches infinity, such that the number of slices is propotional to the inverse of the widths. $\endgroup$
    – Vaelus
    Jul 26, 2019 at 15:33
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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: enter image description here

We can do this in 2 simple steps:

step 1: just moving a triangle step 1: just moving a triangle

step 2: moving a triangle again step 2: moving a triangle again

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    $\begingroup$ @JahanClaes I disagree. This proof, like the "easy to see" first example in the question, needs no information about the area of the triangle beyond the fact that the area is unchanged when you translate the triangle. In the answer's last figure, all that's needed is that triangles ABC and DCE have the same area (nothing about ADC, which would indeed be circular reasoning). $\endgroup$
    – Andreas Blass
    Jul 26, 2019 at 0:55
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    $\begingroup$ I like this one a lot. Somehow induction is more satisfying to visualize than the other visual "proofs". $\endgroup$
    – Vaelus
    Jul 26, 2019 at 15:37
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    $\begingroup$ @AndreasBlass Withdrawn! I didn't think about the figure long enough to realize that in figure 2 the triangles were still congruent. Now this one is my favorite. $\endgroup$
    – Jahan Claes
    Jul 26, 2019 at 16:09
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    $\begingroup$ This doesn't work so nicely if the $x$-coordinate of $E$ in the first diagram is irrational. $\endgroup$
    – aschepler
    Jul 27, 2019 at 10:49
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    $\begingroup$ @aschepler: In that case it doesn't get you to a square, but it still gets you to a parallelogram with an internal altitude: that is, the "easy to see" diagram in the question. $\endgroup$
    – Micah
    Jul 28, 2019 at 0:52
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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.

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Easy visualization for the first example: (Quick and dirty, don't have good art software on this computer.)

enter image description here

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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.)

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    $\begingroup$ It's not clear to me OP doesn't see the figures have equal bases and heights. Presumably OP sees this, and understands the formula therefore dictates they have the same area, but doesn't understand why they ought to have the same area besides taking a formula for granted. $\endgroup$
    – runway44
    Jul 24, 2019 at 22:46
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    $\begingroup$ @runway44 could be: although it could have been more clearly stated that a proof independent of the formula was requested. $\endgroup$
    – rschwieb
    Jul 25, 2019 at 1:05
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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.

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A kite is flying into the sky!

![enter image description here

In fact, it is what you and Stijn already observed. This drawing summarizes it.

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