4 ms·

How is it trivial?

• camjw 3y ago
  Commented this on another thread: Okay so take the triangle made by taking the diagonal of the unit square. This has side lengths 1, 1, and c and has area 1/2. Now, take four of these and arrange them in a square with the side length being c. It would be easier to draw this... basically you stick the right angles in the center. If this isn't clear I can draw a diagram. Anyway, you just made a square with side length c but since its made of four of those original triangles we know that the area of it is 4 * (1/2) = c^2 so c^2 = 2. EDIT: made an excalidraw to explain this construction - maybe helpful https://excalidraw.com/#room=2298a8fd232d5f58e8ca,HmUwSqOt6J https://excalidraw.com/#room=2298a8fd232d5f58e8ca,HmUwSqOt6J...
  • chengiz 3y ago
    You're using geoemetrical construction not dissimilar from proving the theorem for a != b. So it's not in the spirit of this new method. No one disputes there are easier methods to prove the theorem.
• tzs 3y ago
  Note that when a = b we have an equilateral triangle, with area a^2/2. Draw a line from the 90 degree angle to side c, bisecting the 90 degree angle into two 45 degree angles. This divides the original triangle into two smaller triangles. From the fact that the sum of the interior angles of a triangle is 180 degrees, it is not hard to see that the two smaller triangles are both equilateral, with sides of a, c/2, and c/2, and the angle between their two c/2 sides is 90 degrees. That gives c^2/8 for the area of each of the smaller triangles, or c^2/4 for the area of the original triangle which we know to be a^2/2. So c^2/4 = a^2/2 or c^2 = 2 a^2 = a^ + a^2.
  • ipince 3y ago
    Ok, but those triangles aren't equilateral but isosceles :)