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Question Number 116770 by bounhome last updated on 06/Oct/20

how to prove the Pythagorean theorem(c^2 =a^2 +b^2 )

$${how}\:{to}\:{prove}\:{the}\:{Pythagorean}\:{theorem}\left({c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$

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

α+β+γ=180°∧γ=90° ⇒ β=90°−α (a/(sin α))=(b/(sin β))=(c/(sin γ)) (a/(sin α))=(b/(sin (90°−α)))=(c/(sin 90°)) (a/(sin α))=(b/(cos α))=c ⇒ a=csin α ∧ b=ccos α a^2 +b^2 =c^2 sin^2 α +c^2 cos^2 α = =c^2 (sin^2 α +cos^2 α)=c^2

$$\alpha+\beta+\gamma=\mathrm{180}°\wedge\gamma=\mathrm{90}°\:\Rightarrow\:\beta=\mathrm{90}°−\alpha \\ $$$$\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{sin}\:\beta}=\frac{{c}}{\mathrm{sin}\:\gamma} \\ $$$$\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{sin}\:\left(\mathrm{90}°−\alpha\right)}=\frac{{c}}{\mathrm{sin}\:\mathrm{90}°} \\ $$$$\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{cos}\:\alpha}={c} \\ $$$$\Rightarrow\:{a}={c}\mathrm{sin}\:\alpha\:\wedge\:{b}={c}\mathrm{cos}\:\alpha \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\alpha\:+{c}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\alpha\:= \\ $$$$={c}^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \:\alpha\:+\mathrm{cos}^{\mathrm{2}} \:\alpha\right)={c}^{\mathrm{2}} \\ $$

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