proof of Pythagorean theorem

This is a geometrical proof of Pythagorean theorem. We begin with our triangle:

$$,(0,0);(20,0)**\mathrm{@}-;(20,10)**\mathrm{@}-;(0,0)**\mathrm{@}-,(10,-2)*a,(23,6)*b,(10,7)*c$$

Now we use the hypotenuse as one side of a square:

$$,(0,0);(20,0)**\mathrm{@}-;(20,10)**\mathrm{@}-;(0,0)**\mathrm{@}-;(-10,20)**\mathrm{@}-;(10,30)**\mathrm{@}-;(20,10)**\mathrm{@}-,(10,-2)*a,(23,6)*b,(10,7)*c$$

and draw in four more identical triangles

$$,(0,0);(20,0)**\mathrm{@}-;(20,10)**\mathrm{@}-;(0,0)**\mathrm{@}-;(-10,20)**\mathrm{@}-;(10,30)**\mathrm{@}-;(20,10)**\mathrm{@}-;(20,30)**\mathrm{@}-;(-10,30)**\mathrm{@}-;(-10,0)**\mathrm{@}-;(0,0)**\mathrm{@}-,(10,-2)*a,(23,6)*b,(10,7)*c$$

Now for the proof. We have a large square, with each side of length $a+b$, which is subdivided into one smaller square and four triangles. The area of the large square must be equal to the combined area of the shapes it is made out of, so we have

${\left(a+b\right)}^2$ $=$ $c^2+4\left({\displaystyle \frac{1}{2}}ab\right)$

$a^2+b^2+2ab$ $=$ $c^2+2ab$

$a^2+b^2$ $=$ $c^2$ (1)

Title	proof of Pythagorean theorem
Canonical name	ProofOfPythagoreanTheorem
Date of creation	2013-03-22 11:56:36
Last modified on	2013-03-22 11:56:36
Owner	drini (3)
Last modified by	drini (3)
Numerical id	8
Author	drini (3)
Entry type	Proof
Classification	msc 51-00
Related topic	PythagorasTheorem