Proving Pythagoras’ theorem
http://topdrawer.aamt.edu.au/Geometric-reasoning/Good-teaching/Writing-aproof/Proving-Pythagoras-theorem/Dissected-proof
In any right angled triangle, the square on the hypotenuse is equal to the sum of the
squares on the other two sides.
Aim: To prove c2 = a2 + b2
A
A
x
D
c
c
b
y
a
C
A
b
B
a
C
B
x
D
D
b
y
C
C
a
B
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page 1 of 2
c
a
b
=
=
b CD y
In ΔABC and ΔADC
(matching sides of similar triangles)
(both 90° given)
∠ACB = ∠ADC
∴ c2 = a2 + b2
∴ ΔABC ||| ΔACD
is common
∠A
∴ ΔABC ||| ΔCBD
∴
AB BC AC
=
=
AC CD AD
∴ a2 = cx
(AAA)
∴
c b
=
b y
(both 90° given)
∴ b2 = cy
∠B
(AAA)
Now a2 + b2 = cx + cy
∠ACB = ∠BDC
= c(c)
In ΔABC and ΔBDC
(matching sides of similar triangles)
= c(x + y)
is common
= c2
c a
b
= =
a x CD
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page 2 of 2
∴
AB BC AC
=
=
CB BD CD