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openai:o4-mini-2025-04-16 via @hdevalence.bsky.social
Thinking Summary

**Generating Pythagorean Theorem Proof** The user wants me to generate a proof of the Pythagorean theorem and explain it clearly. I’ll use the area rearrangement proof, which visualizes squares on the sides of a right triangle showing that the area of the larger square equals the sum of the areas of the smaller squares. I can describe the setup with a triangle ABC, and drop an altitude from C to the hypotenuse AB, utilizing similar triangles to derive relationships. I’ll format everything in markdown using KaTeX syntax.

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=bAC = b, BC=aBC = a, AB=cAB = c.
  • CDCD is the altitude, splitting ABAB into segments ADAD and DBDB.

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

Step 2: Write ratios from similarity

From △ACD ∼ △ABC:

  • ACAB=ADAC\displaystyle\frac{AC}{AB} = \frac{AD}{AC}
  • Hence AC2=ABADAC^2 = AB\cdot AD, i.e. b2=cADb^2 = c\cdot AD.

From △CBD ∼ △ABC:

  • BCAB=BDBC\displaystyle\frac{BC}{AB} = \frac{BD}{BC}
  • Hence BC2=ABBDBC^2 = AB\cdot BD, i.e. a2=cBDa^2 = c\cdot BD.

Step 3: Add the two equations

Adding  b2=cADb^2 = c\cdot AD
a2=cBDa^2 = c\cdot BD
gives
a2+b2=c(AD+BD).a^2 + b^2 = c\cdot(AD + BD).

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

a2+b2=cc=c2.a^2 + b^2 = c\cdot c = c^2.

That is,

a2+b2=c2,a^2 + b^2 = c^2,

which is precisely the Pythagorean theorem.


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 b2b^2 and a2a^2 each equaling a part of c2c^2.
  • Summing recovers the full square on the hypotenuse.

This proof beautifully ties together geometry (altitude, similarity) and algebra (multiplying out ratios) to establish a2+b2=c2a^2 + b^2 = c^2\,.

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