So, to be clear, he’s only talking about a hexagon, which is a polygon, not a polygon which may or may not be a hexagon. I’m on dangerous ground here cuz I never did geometry well. But I can use Illustrator, or even a pencil and a ruler, to create a six sided polygon with no equilateral triangles whatsoever. Please abuse me of this notion.
No need for abuse 
“Hexagon” is just another word for six-sided polygon. It does not contradict anything that not every single one can be dissected into equilateral triangles. You could have a very small angle somewhere, for example.
Yeah, you got it. And if you use a square as your “central cookie cutter,” and you stamp four of them around in a circle, you’ll end up creating an eight-sided shape. It will look like 4 sides, but that’s just because each of the eight sides is colinear with another.
That is, when thinking about a baseball diamond, you can imagine a small square with opposite corners at the pitcher’s rubber and home plate. The other two would be halfway up the first base line and the third base line. A square is the 4-sided cousin of the equilateral triangle.
Four such squares exist around the pitcher, one facing each base. It just so happens that the home-to-halfway-to-first segment is colinear with the from-halfway-there-to-the first-base-bag segment. Likewise, the second-base-to-shortstop is colinear with the shortstop-to-third segment. If you think of it that way, the original four-sided cookie cutter shape makes a bigger one with eight sides.
I wonder if this works for all shapes.
If you are asking about dividing shapes into several similar pieces, here are a few: mine
Still nobody has mentioned that, treating the Earth as a sphere, the equator and meridians through London UK (0°), Dhaka BGD (90°E), Savusavu Fiji(180°) and Memphis TN (90°W) divide the Northern and Southern Hemispheres each into FOUR (not six) equilateral triangles.
Yeah; I know I have a reputation of being a trouble-maker. 
Correct in your general understanding, but it turns out that the title and the actual question refer to two different things. The title talks about “a six-sided polygon”, which is just a closed area limited by six fragments of straight line. “Hexagon” is the same, with more Greek (hexa = 6). But then the actual question refers to a regular hexagon, which is the one where the 6 line fragments are the same length and the 6 angles between them are all the same.
A rhombus, a rectangle and a square are all “tetragons” or “four-sided polygons”. But only the square is regular.
Except that, for some reason, we use the term “quadrilateral” instead of “tetragon”. But yeah, the point is the same.
Yup. As mentioned before, this depends on Euclid’s Fifth Postulate, which is only true in flat spaces, and the surface of the Earth is not a flat space.
IMHO, I think all the answers above have missed the mark.
Question: Why do equilateral triangles fit perfectly into a hexagon?
Short answer: Because 1/2 - 1/3 = 1/6.
Easier to understand answer: If you walk straight ahead 10 feet, make a u-turn, pass your starting point and go another 10 feet, make another u-turn, then go back to the precise spot you started from, you just traced two 20-foot line segments on top of each other, with two 1/2 turns. Now try it with three turns instead of two. Walk straight ahead ten feet and make a turn to the right. How much should the turn be? Well, you’re going to make three of them and you want them to be equal, so each turn has to be 1/3. Walk straight ahead 20 feet and make another 1/3 turn. Walk 20 more feet and make another 1/3 turn. Walk 10 more feet back to your precise starting point. You’ve traced an equilateral triangle where each side is 20 feet long. What is the measure of each angle of the triangle? At each corner, you could have made a 1/2 turn to retrace your steps but instead you made a 1/3 turn. What’s the difference between 1/2 and 1/3 ? 1/2 - 1/3 = 1/6. If each corner is 1/6, how many can you fit at the same apex? Six of them. If you look at the outside edges, you just made a regular hexagon. If you prefer to use 360º to describe a full turn, then it goes like this: A u-turn is 180º. When you walk the triangle, you make three 120º turns. Each one forms an angle of 180º-120º=60º. And you can fit exactly six of those into 360º.
Advanced answer: This works with any number of sides n>2. Walk an n-sided regular polygon and each corner requires a turn of 1/n (or 360º/n) and forms angles of 1/2 - 1/n (or 180º - 360º/n). How many you can fit at the same apex is 1 / (1/2 - 1/n) or 360º/(180º-360º/n). When n=3 you get triangles making 60º angles and six of them will form a hexagon. When n=4 you get squares making 90º angles and four of them will form a bigger square because 1/2 - 1/4 = 1/4. When n=5 you get 1/2 - 1/5 = 3/10 which means you can fit 3.333333 of them at the same apex and they don’t form anything. When n=6 you get hexagons and 1/2 - 1/6 = 1/3 so three of them will fit at the same apex. They don’t form a regular polygon but it’s a good start at making a honeycomb. The only values for n>2 where 1 / (1/2 - 1/n) is a whole number are n=3 and n=6.
Of course, all this assumes a flat surface and not, for example, the surface of a sphere.
I meant to say n=3, n=4, and n=6.
Six copies of any triangle can be always be arranged around a point with no gaps (for the same reason three copies can always be arranged to form a straight angle: the sum of the angles in a triangle is 180 degrees).
For values of “always” which include “in flat space”.
sbunny’s explanation is really good, but I think there’s one bit of clarification. The goal is to arrive back at your original location, and be facing the same way, so you could do it again. So the “u-turns” are not actually “u-turns,” but turning in place half-way around. And the last turn is always made when you’re at the starting point, getting ready to trace the shape again.
I’ll also add that, when you put the triangles together, you are putting together one angle, and two sides. So that two of the sides disappear. And since it took six triangles, that means 6 sides are left, a hexagon. And since the angles of the triangles were all the same, the angles of the hexagon are the same: a regular hexagon.
Going beyond cutting a hexagon into six triangles, as a fun jigsaw puzzle for your game room, try assembling equilateral triangles of sides 2, 2, 3, 5, 5, 7, 7, 8, 8, 9, 11, 11, 12, 19, and 20 into a big equilateral triangle. Also try dissecting a perfect square into 45°-60°-75° triangles. The point is that these kinds of problems always have elementary but interesting mathematics behind them, and often result in practical and attractive puzzles.
What’s an .nb file? I’m guessing that stands for “notebook”, and coming from Wolfram, it’s probably something Mathematica-related, but most users aren’t going to be able to open that.
Hmm; you are right. Here is the correct link:
[Could someone please edit and replace the bad link with the good one?]
OK, that’s better than I was going to do. I was aiming for a fractal infinite-triangle solution, with triangles around the edge and another square in the middle (which of course would then be tiled the same way).
That works, of course!
I don’t actually know the solution with the minimal number of triangles (one with 46 is given in one of the links), but with fewer than 50 triangles a brute-force computer search should resolve this. I may even eventually get around to it…