Question Number 14661 by ajfour last updated on 03/Jun/17
Commented by ajfour last updated on 03/Jun/17
$$\boldsymbol{{Prove}}\:\boldsymbol{{Pythagoras}}\:\boldsymbol{{theorem}} \\ $$$${with}\:{the}\:{aid}\:{of}\:{image}\:{above}\:. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 03/Jun/17
$${h}^{\mathrm{2}} =\mathrm{4}×\frac{\mathrm{1}}{\mathrm{2}}{bp}+\left({b}−{p}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} +{p}^{\mathrm{2}} \:.\blacksquare \\ $$
Commented by ajfour last updated on 03/Jun/17
$${how}\:{nice},\:{you}\:{wer}\boldsymbol{{e}}\:\boldsymbol{{q}}{uick}\:! \\ $$