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. ]

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Proof 1: Rearrangement (Bhāskara’s proof)

1. Start with a large square of side length (a+b). Its area is
   [ (a+b)^2. ]

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):

   ┌─────────────────────────┐
   │▢──▶                    ◀──▢│
   │ │                      │ │
   │ ▼                      ▲ │
   │                        │ │
   │ ▲                      ▼ │
   │ │                      │ │
   │◀──                    ──▶│
   └─────────────────────────┘
   

3. 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. ]

4. Expand and simplify: [ a^2 + 2ab + b^2 ;=; 2ab + c^2 \quad\Longrightarrow\quad a^2 + b^2 = c^2. ]

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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

1. 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.

2. 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}. ]

3. 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}. ]

4. Since (AD + BD = AB),
   [ \frac{b^2}{c} + \frac{a^2}{c} = c \quad\Longrightarrow\quad a^2 + b^2 = c^2. ]

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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.} ]