I reached a contradiction while proving pythagorean theorem, but I don't know what I did wrong?

Question

By relocating the two triangles in the bottom to the top, I created the rectangle with sides $a$ and $b+b=2b$. Then, doesn't this imply that $c^2=2ab$? I am thinking that the area of four triangles add up to $c^2$ and using this fact to prove the pythagorean theorem. I know $c^2=a^2+b^2$, but this makes perfect sense to me.

Area of your paralellogram is $c^2\sin2A$. – ziggurism Dec 21 '16 at 04:11 — ziggurism, Dec 21 '16 at 04:11
The upper figure is a square only when a=b. – DanielWainfleet Dec 21 '16 at 05:52 — DanielWainfleet, Dec 21 '16 at 05:52

score 3 · Accepted Answer · answered Dec 21 '16 at 03:55

3

The quadrilateral at the top is not a square, so its area is not $c^2$.

answered Dec 21 '16 at 03:55

angryavian

89,882

I reached a contradiction while proving pythagorean theorem, but I don't know what I did wrong?

1 Answers1