Pythagore:exercice\ utilisant \ la\ réciproque\ du\ théorème.

A

B

C

28

45

53

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

Le\ grand\ côté\ du\ triangle\ ABC\ est\ [BC]\ donc\ si\ ce\ triangle\ est\ rectangle\, ce\ sera\ en\ A.

Calculons\ séparément:

AB²+AC²\
=28²+45²\
=784+2025\
=2809

et

BC²\
=53²\
=2809

On\ constate\ que:

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

AB²+AC²=BC².

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

la\ \textbf{\textit{réciproque}}\ du\ théorème\ de\ Pythagore.

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

AB²+AC²=BC².

\textbf\textit{{le\ triangle\ ABC\ est\ rectangle\ en\ A}}.

\textbf {Exercice}

\textbf {Corrigé}

MathsEnClair.com

A = (-83.24640234252838, 25.81485008912665)
B = Point(Circle[A, 28])
a = Segment(A, B)
c = Circle(A, 45)
d = Circle(B, 53)
C = Intersect(c, d, 1)
b = Segment(A, C)
e = Segment(C, B)
D = Midpoint(A, B)
E = Midpoint(A, C)
F = Midpoint(B, C)
etape = 0 (slider from 0 to 13, step 1)