Is my logic correct below to prove the Pythagoras Theorem? Thanks.

Area Rectangle R \begin{align*} R &= WL\\ &=(2a+b)(2b+a)\\ &=4ab+2a^2+2b^2+ab\\ &=5ab+2a^2+2b^2 \end{align*}

Total Area Yellow Triangles T \begin{align*} T &= 10(\frac{ab}{2})\\ &=5ab \end{align*}

Calculate Area $c^2$

\begin{align*} c^2 &= R-T-a^2-b^2 \\ &=5ab+2a^2+2b^2-5ab-a^2-b^2\\ &=a^2+b^2 \end{align*}


Proof by rearrangement


  • $\begingroup$ See also Bhaskara's proof. $\endgroup$ – Oscar Lanzi Nov 30 '20 at 0:43
  • $\begingroup$ Bhaskara Proof is from cut-the-knot.org/pythagoras Proof #3 Is my proof new or already discovered? Thanks! $\endgroup$ – vengy Nov 30 '20 at 0:48
  • $\begingroup$ For something as worked over as the Pythagorean Theorem, assume it's already discovered. A very extensive literature search would be needed to prove otherwise. $\endgroup$ – Oscar Lanzi Nov 30 '20 at 0:55
  • $\begingroup$ True. It seems similar to Proof #87 from cut-the-knot.org/pythagoras $\endgroup$ – vengy Nov 30 '20 at 3:02

Yes. This is correct provided you have shown that all yellow triangles are congruent.

  • $\begingroup$ Thanks! Google'd what congruent means - "they will have exactly the same three sides and exactly the same three angles". Is it easy to show congruency above? I just re-orientated the triangle abc, which should keep the sides/angles congruent. $\endgroup$ – vengy Nov 29 '20 at 20:46
  • $\begingroup$ SAS (side-angle-side) congruency is sufficient. They each have two smaller sides $a$ and $b$ and one included right angle. $\endgroup$ – cosmo5 Nov 29 '20 at 20:47
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    $\begingroup$ Yes ofcourse! It is neat. +1 For more visual proofs like this, also see cut-the-knot.org/pythagoras $\endgroup$ – cosmo5 Nov 29 '20 at 20:58
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    $\begingroup$ Wow...thanks for verifying cosmo5! $\endgroup$ – vengy Nov 29 '20 at 21:01
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    $\begingroup$ Thanks! It's a non-algebraic proof. Got the idea from this simple animation: en.wikipedia.org/wiki/… $\endgroup$ – vengy Dec 1 '20 at 17:53

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