Pythagoras’ Theorem (Similar Triangles Proof)
Y9
C
This proof is credited to Adrian Marie Legendre. A famous
French mathematician, born Toulouse 1752, died Paris 1833.
a
b
Here we have a right angled triangle ABC. We drop a
perpendicular from C to AB to Make X.
A
B
c
The distance AX is x. Therefore XB is c - x.
C
∆s ABC and CBX are similar.
∠B is common, both have 1 right angle.
∴ the third angles must be the equal (angle sum of triangle).
∆s ABC and ACX are similar.
∠A is common, both have 1 right angle.
∴ the third angles must be the equal (angle sum of triangle).
a
b
c -x
x
A
B
X
∆s ABC, CBX, and ACX are similar.
Since corresponding parts of similar triangles are proportional,
In ∆s ABC and ACX b = c x
b
b² = cx.
In ∆s ABC and CBX a = c (c - x) a
a² = c² - cx
Substituting b² for cx, we get c² = a² + b²
c² = a² + cx
(Pythagoras’ Theorem)
QED
Using this method, prove Pythagoras’ Theorem for these triangles.
1).
F
W
d
e
x
f -y
y
D
2).
Y
E
f
3).
v
T
V
v
X
R
4).
U
p
q
u
Z
w
t
V
P
r
Q
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