Question Number 116770 by bounhome last updated on 06/Oct/20
$${how}\:{to}\:{prove}\:{the}\:{Pythagorean}\:{theorem}\left({c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$
Commented by $@y@m last updated on 06/Oct/20
https://www.math-only-math.com/proof-of-pythagorean-theorem.html#:~:text=The%20proof%20of%20Pythagorean%20Theorem,of%20c%20(c2).
Answered by MJS_new last updated on 06/Oct/20
$$\alpha+\beta+\gamma=\mathrm{180}°\wedge\gamma=\mathrm{90}°\:\Rightarrow\:\beta=\mathrm{90}°−\alpha \\ $$$$\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{sin}\:\beta}=\frac{{c}}{\mathrm{sin}\:\gamma} \\ $$$$\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{sin}\:\left(\mathrm{90}°−\alpha\right)}=\frac{{c}}{\mathrm{sin}\:\mathrm{90}°} \\ $$$$\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{cos}\:\alpha}={c} \\ $$$$\Rightarrow\:{a}={c}\mathrm{sin}\:\alpha\:\wedge\:{b}={c}\mathrm{cos}\:\alpha \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\alpha\:+{c}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\alpha\:= \\ $$$$={c}^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \:\alpha\:+\mathrm{cos}^{\mathrm{2}} \:\alpha\right)={c}^{\mathrm{2}} \\ $$