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Here’s a classical proof of the Pythagorean theorem using similar triangles (often attributed to Euclid I.47):

Setup

1. Let △ABC be a right triangle with right angle at C.
2. Let AB be the hypotenuse, and drop the altitude from C onto AB, meeting it at D.

      A
      /|
     / |
   c/  | b
   /   |
  /____|
 C   a  B

• $AC = b$, $BC = a$, $AB = c$.
• $CD$ is the altitude, splitting $AB$ into segments $AD$ and $DB$.

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Step 1: Identify similar triangles

Because each of the angles at A, B, C is shared or complementary, we have three pairwise-similar right triangles:

• △ABC ∼ △ACD (they share ∠A and both have a right angle)
• △ABC ∼ △CBD (they share ∠B and both have a right angle)

In particular:

• △ACD ∼ △ABC
• △CBD ∼ △ABC

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Step 2: Write ratios from similarity

From △ACD ∼ △ABC:

• $\displaystyle\frac{AC}{AB} = \frac{AD}{AC}$
• Hence $AC^2 = AB\cdot AD$, i.e. $b^2 = c\cdot AD$.

From △CBD ∼ △ABC:

• $\displaystyle\frac{BC}{AB} = \frac{BD}{BC}$
• Hence $BC^2 = AB\cdot BD$, i.e. $a^2 = c\cdot BD$.

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Step 3: Add the two equations

Adding  $b^2 = c\cdot AD$
 $a^2 = c\cdot BD$
gives
$a^2 + b^2 = c\cdot(AD + BD).$

But $AD + BD = AB = c$, so

$a^2 + b^2 = c\cdot c = c^2.$

That is,

$a^2 + b^2 = c^2,$

which is precisely the Pythagorean theorem.

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Why it works

• Dropping the altitude creates two smaller right triangles each similar to the original.
• Similarity gives proportionality of sides, which translates into the two “partial” squares $b^2$ and $a^2$ each equaling a part of $c^2$.
• Summing recovers the full square on the hypotenuse.

This proof beautifully ties together geometry (altitude, similarity) and algebra (multiplying out ratios) to establish $a^2 + b^2 = c^2\,$.