proof of Pythagorean theorem

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

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

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 fourtriangles. The area of the large square must be equal to the combinedarea 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)