Question 415334
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In order for the 4 triangles to form a square, the triangles must be isosceles right triangles.  That is to say *[tex \Large a\ =\ b] and a hypotenuse of one of the right triangles forms one of the sides of the square.


The area of the square is then *[tex \Large c^2]


But the area of one of the triangles is *[tex \Large \frac{ab}{2}], hence the square, being composed of 4 triangles has an area of *[tex \Large 2ab]


Since we have established that *[tex \Large a\ =\ b], *[tex \Large 2ab] can be written *[tex \Large 2a^2] which is to say *[tex \Large a^2\ +\ a^2].  Furthermore, again since *[tex \Large a\ =\ b], *[tex \Large a^2\ +\ a^2\ =\ a^2\ +\ b^2].


Since both *[tex \Large a^2\ +\ b^2] and *[tex \Large c^2] are expressions for the same area, namely square EFGH, we can equate the two expressions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c^2\ =\ a^2\ +\ b^2]


Which is the Pythagorean Theorem.


Unfortunately, this logic does not strictly prove Pythagora's conjecture.  That is because we created a situation at the beginning that required the triangles to be isosceles right triangles.  Hence, this proof is only strictly valid for isosceles right triangles and not right triangles in general.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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