can you generate and explain a proof of the pythagorean theorem
Thread (2 posts)
Here is a classic statement and two proofs of the Pythagorean Theorem.
Theorem.
In any right triangle with legs of lengths (a) and (b) and hypotenuse of length (c),
[
a^2 + b^2 = c^2.
]
Proof 1: Rearrangement (Bhāskara’s proof)
-
Start with a large square of side length (a+b). Its area is
[ (a+b)^2. ] -
Inside it place four copies of the right triangle (legs (a,b), hypotenuse (c)) so that they form a smaller tilted square in the center whose side is (c). The picture looks like this (▢ = triangle):
┌─────────────────────────┐ │▢──▶ ◀──▢│ │ │ │ │ │ ▼ ▲ │ │ │ │ │ ▲ ▼ │ │ │ │ │ │◀── ──▶│ └─────────────────────────┘
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The area of the large square equals the area of the four triangles plus the area of the central square: [ (a+b)^2 ;=; 4\cdot\bigl(\tfrac12,a,b\bigr);+;c^2. ]
-
Expand and simplify: [ a^2 + 2ab + b^2 ;=; 2ab + c^2 \quad\Longrightarrow\quad a^2 + b^2 = c^2. ]
Proof 2: Similar triangles
Consider right triangle (ABC) with right angle at (C). Let the altitude from (C) meet hypotenuse (AB) at (D).
A
•
│\
b│ \ c
│ \
•───• B
C a
-
Triangles (ABC), (ACD), and (CBD) are all similar:
- (\triangle ABC \sim \triangle ACD) because they share angle at (A) and both are right at (C)/(D).
- (\triangle ABC \sim \triangle CBD) similarly.
-
From similarity (\triangle ABC \sim \triangle ACD): [ \frac{AB}{AC} = \frac{AC}{AD} \quad\Longrightarrow\quad \frac{c}{b} = \frac{b}{AD} \quad\Longrightarrow\quad AD = \frac{b^2}{c}. ]
-
From similarity (\triangle ABC \sim \triangle CBD): [ \frac{AB}{BC} = \frac{BC}{BD} \quad\Longrightarrow\quad \frac{c}{a} = \frac{a}{BD} \quad\Longrightarrow\quad BD = \frac{a^2}{c}. ]
-
Since (AD + BD = AB),
[ \frac{b^2}{c} + \frac{a^2}{c} = c \quad\Longrightarrow\quad a^2 + b^2 = c^2. ]
Both proofs conclude that in a right triangle, the sum of the squares on the legs equals the square on the hypotenuse:
[
\boxed{a^2 + b^2 = c^2.}
]