What is the simplest proof of the pythagorean theorem you know? Maybe enough so to explain it to children. 
 A: I wouldn't call  this a proof, but it's a convincing argument good demonstration of what the theorem means, and I thought it was pretty cool. Hopefully should help with explaining it to kids, as you asked.

A: 'Picture proofs' are particularly elegant. I like this one:

A: I liked Garfield's Proof. It is simple and intuitive.
Here, https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-geometry/cc-8th-pythagorean-proofs/v/garfield-s-proof-of-the-pythagorean-theorem
A: For geometric proof using rearrangment there are few quite good example on the Wikipedia page. A beginner likely will not understand this quite well, because all those triangles in the picture can be confusing, so I recommend to start first with this "counting proof"

Then introduce the rearrangment pictures and animations, and at last introduce the actual formula $a^2 + b^2 = c^2$. It would be nice to lead them and help them to derive the formula by themselves, rather than just writing it on a board.
A: You can cut up a square of sides $a+b$ into $a^2$, $b^2$ and four triangles of sides $a,b,c$.
You can also cut up the same square into a square of side $c$, and the same four triangles.
A: Consider the right figure and ignore the left one!

The area of the square of side $a+b$ is 
\begin{align*}
(a+b)(a+b) 
&= a^2 + ab + ab + b^2\\
&= a^2 + 2ab + b^2
\end{align*}
It is equal to the sum of 4 triangles and 1 square. This sum is
\begin{gather*}
4\times \frac{ab}{2} + c^2\\
2ab+ c^2
\end{gather*}
By equating both, we have
\begin{align*}
a^2 + 2ab + b^2
&= 2ab +c^2\\
a^2+b^2=c^2
\end{align*}
Now consider a single triangle so you have proven that $a^2+b^2=c^2$. Done!
A: Every time you walk on a floor that is tiled like this, you are walking on a proof of the Pythagorean theorem.

EDIT: Due to popular demand, I have added the grid in red on the right, with some triangle legs in blue.
If you consider say the upper left corner of every small square, you can see that these points lie on a slightly diagonal periodic grid.
Each square in this grid is $c \times c$.
Then you can choose either of two easy proofs:

(1) The $c \times c$ square is clearly a rearrangement of one $b \times b$ square and one $a \times a$ square.
or
(2) Each period of the periodic pattern covers what area of floor?  On the one hand there is one $a \times a$ square and one $b \times b$ square per period, while on the other hand there is one $c \times c$ square per period.

(The second proof is my favorite, since unlike most proofs it requires neither dissection nor algebra.  Once you see that the tiling's periodicity is $c \times c$, you are done!)
A: I think that one of the simplest proofs is that attributed to US President James Abram Garfield.
The Cut the Knot page on the Pythagorean lists it as Proof #5. 


This proof, discovered by President J. A. Garfield in 1876 [Pappas],
  is a variation on the previous one. But this time we draw no squares
  at all. The key now is the formula for the area of a trapezoid - half
  sum of the bases times the altitude - $(a + b)/2·(a + b)$. Looking at
  the picture another way, this also can be computed as the sum of areas
  of the three triangles - $ab/2 + ab/2 + c·c/2$. As before,
  simplifications yield a² + b² = c². 

For more of this story, see: A Mathematician for President, Let's Play Math.

PS
I just noticed that this proof was already given as an answer. I'll leave my response here as it contains more details.
PPS
My next favourite proof is probably the one based on a construction of similar triangles (Proofs #6 and #7). I think it cuts to the heart of the idea as seen by Euclid.
Then probably the simple picture proofs #3, #4, #9 - the latter already being posted as an answer to this question.
The original Euclidean proof constructed from the "Bride's Chair" always seemed a little convoluted to me.
A: What I will say will be the worst in terms of beauty but you need to ask yourself what is a square triangle. How is it defined ?
Well, the answer is that 2 vectors are square when their scalar product is 0, which leads (after 1 or 2 lines) to the Pythagore Theorem
Edit :
Vectors are A and B ; C = A - B to "complete the triangle"
In the general case, ||C||2 = (A-B).(A-B) = ||A||2 + ||B||2 - 2 A.B
We see that ||C||2 = ||A||2 + ||B||2 if and only if A.B == 0 (i.e are square by definition of being square)
