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Théorème\ de\ Pythagore\ Preuve\ par\ le\ calcul\ littéral\ (\ trapèze\ rectangle)

Rappel:\\mathcal {A}_{trapèze}=\frac{petite\ base+ grande\ base}{2}\times h.

1°)\ Quelle\ est\ la\ nature\ du\ quadrilatère\ MNPQ\ ?\ En\ déduire\ que:\mathcal {A}_{MNPQ}=\frac{(a+b)²}{2}

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2°) \ Montrer\ que\ le\ triangle\ QRP\ est\ rectangle\ en\ R.\ En\ déduire\ l'aire\ du\ triangle\ QRP\ en\ fonction\ de\ c.

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3°)\ En\ calculant\ de\ deux\ façons\ différentes\ \mathcal{A}_{MNPQ},\ montre\ que\ :a²+b²=c².

On\ sait\ que\ (MQ)\perp(MN)\ et\ (NP)\perp(MN).\ On\ utilise:“si\ deux\ droites\ sont\ perpendiculaires\ à\ une\ même\ droite,\
alors\ elles\ sont\ parallèles“.\ On\ en\ déduit\ que\ (MQ)//(NP), puis\ que\ MNPQ\ est\ un\ trapèze\  rectangle\ de\ bases\ [MQ]\ et\ [NP].\
La\ formule\ de\ l'aire\ d'un\; trapèze\ donne:\mathcal {A}_{MNPQ}=\frac{MQ+NP}{2} \times MN=\frac{a+b}{2}\times(a+b)=\frac {(a+b)²}{2}.\

Posons:\alpha=\widehat{MRQ}=\widehat{NPR}, et\ \beta=\widehat{MQR}=\widehat {NRP}:\alpha +\beta=90°. \ On\ a:\widehat {NRM}=\widehat{NRP}+\widehat{PRQ}+\widehat{QRM}.\
En\ remarquant\ que\ \widehat{NRM}\ est\ un\ angle\ plat, on\ ontient :180°=\beta+\widehat{PRQ}+\alpha \Leftrightarrow \widehat{PRQ}=180°-(\alpha+\beta).\ 
Or, \alpha+\beta=90°, donc:\widehat{PRQ}=180°-90°=90°,et\ par\ conséquent\ le\ traingl\ QRP\ est\:\ rectangle\ en\ R.\
\mathcal {A}_{QRP}=\frac{1}{2}\times produit\ des\ côtés\ de\ l'angle\ droit=\frac{1}{2}\times RP\times RQ=\frac {1}{2}\times c\times c =\frac {c²}{2}.

On\ a:\mathcal {A}_{MNPQ}= \mathcal {A}_{MRQ}+ \mathcal {A}_{QRP}+ \mathcal {A}_{RNP}=\frac {a\times b}{2}+\frac {c²}{2}+\frac {a\times b}{2}=\frac {c²+2ab}{2}.\ Or, on\ vu\ en\ 1°) \ que:\mathcal {A}_{MNPQ}=\frac{(a+b)²}{2}.\
Donc:\frac{(a+b)²}{2}=\frac {c²+2ab}{2} \Leftrightarrow \frac {(a+b)(a+b)}{2}=\frac {c²+2ab}{2} \Leftrightarrow \frac{a²+ab+ab+b²}{2} \textcolor{red}{\times 2}=\frac {c²+2ab}{2}\textcolor{red}{\times 2}
\Leftrightarrow a²+2ab+b²=c²+2ab\
\Leftrightarrow a²+2ab+b²-\textcolor {red}{2ab}=c²+2ab-\textcolor{red}{2ab}\Leftrightarrow {a²+b²=c².

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3°)

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MathsEnClair.com

a = 1.8 (slider from 0.5 to 5, step 0.1)
b = 3.2 (slider from 0.5 to 5, step 0.1)
M = (-3.44, 2.06)
R = Point(Circle[M, abs(b)])
c = Segment(M, R)
d = Ray(M, R)
e = Circle(R, a)
N = Intersect(e, d, 2)
f = OrthogonalLine(M, c)
g = OrthogonalLine(N, d)
h = Circle(M, a)
Q = Intersect(h, f, 2)
k = Circle(N, b)
P = Intersect(k, g, 2)
i = Segment(P, Q)
j = Segment(Q, R)
l = Segment(R, P)
α = Angle(R, M, Q)
β = Angle(P, N, R)
m = Segment(Q, M)
n = Segment(N, P)
poly1, d_1, a_1, e_1 = Polygon(M, R, Q)
poly2, c_1, e_2, b_1 = Polygon(R, N, P)
poly3, c_2, d_2, e_3 = Polygon(Q, R, P)
Q': Q - (0.25, 0)
M': M - (0.25, 0)
A = Midpoint(Q', M')
u = Vector(A, Q')
v = Vector(A, M')
M_2: M - (0, 0.25)
R_2: R - (0, 0.25)
N_2: N - (0, 0.25)
B = Midpoint(M_2, R_2)
C = Midpoint(R_2, N_2)
w = Vector(B, M_2)
p = Vector(B, R_2)
q = Vector(C, R_2)
r = Vector(C, N_2)
N_3: N + (0.25, 0)
P_3: P + (0.25, 0)
D = Midpoint(P_3, N_3)
s = Vector(D, N_3)
t = Vector(D, P_3)
o = false
f_1 = false
g_1 = false
E = Midpoint(Q, R)
F = Midpoint(R, P)