Pythagore:exercice\ utilisant\ la\ forme\ contraposée\ du\ théorème.

Le\ triangle\ EFG\ est \text{-}il\ rectangle\ ?

Le\ grand\ côté\ du\ triangle\ EFG\ est\ [EF]\ donc\ si\ ce\ triangle\ est\ rectangle,ce\ sera\ en\ G.

Calculons\ séparément:

GE²+GF²\
=30²+45²\
=900+2025\
=2925

et

EF²\
=55²\
=3025

On\ constate\ que:

\textcolor{red}{\underline{\textcolor{black}{On\ sait\ que}}}:

GE²+GF² \textcolor{red}{\ne}EF².

\textcolor{red}{\underline{\textcolor{black}{On\ utilise}}}:

la\ \textbf{\textit{forme\ contraposée}}\ du\ théorème\ de\ Pythagore.

\textcolor{red}{\underline{\textcolor{black}{On\ en\ déduit\ que}}}:

GE²+GF² \textcolor{red}{\ne}EF².

le\ triangle\ EFG\ n\text{'}est\ pas\ rectangle\ en\ G, ce\ qui\ était\ la\ seule\ possibilité.

\textbf {Exercice}

\textbf {Corrigé}

MathsEnClair.com

E

F

G

30

45

55

etape = 0 (slider from 0 to 13, step 1)
E = (-40.181056053798365, 22.309531205160255)
F = Point(Circle[E, 55])
a = Segment(E, F)
c = Circle(E, 30)
d = Circle(F, 45)
G = Intersect(c, d, 1)
b = Segment(E, G)
e = Segment(G, F)
A = Midpoint(E, G)
B = Midpoint(G, F)
C = Midpoint(E, F)