re: New Orleans East teens make 'impossible' mathematical discovery unproven for 2,000 years

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FWIW, there are several geometrical proofs of Pythagoras' Theorem. That is the basis of the theorem that trigonometry is drawn from. So it is not circular.

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Now ask GPT Chat to prove Pythagoras theory using geometry.

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prove Pythagoras theory using geometry

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Yes, Pythagoras' Theorem can also be proven using geometry. One way to prove it is to construct squares on each side of a right triangle, and then show that the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides.

Here is a step-by-step proof:

Start with a right triangle with legs of length a and b and hypotenuse of length c.

Construct squares on each side of the triangle, with sides of length a, b, and c respectively.

Divide the square on the hypotenuse into two smaller squares, each with sides of length a and b, by drawing a line from the right angle to the midpoint of the hypotenuse.

The area of the square on the hypotenuse is c^2, and the areas of the two smaller squares are a^2 and b^2.

The two smaller squares on the legs of the triangle can be rearranged to form a larger square with sides of length a+b. The area of this square is (a+b)^2 = a^2 + 2ab + b^2.

The area of the larger square is equal to the sum of the areas of the smaller squares, so we have:

a^2 + b^2 = c^2

This is Pythagoras' Theorem.

Therefore, Pythagoras' Theorem can be proven using geometry.

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The proof you posted uses circular logic.

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prove the Pythagorean theorem from trigonometry without resorting to circular logic

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Certainly! Here is a proof of the Pythagorean Theorem from trigonometry that does not use circular logic:

Consider a right triangle with sides a, b, and c, where c is the hypotenuse, and let angle A be the angle opposite side a.

Draw a perpendicular line from angle A to side c, dividing the triangle into two smaller right triangles with sides h, a, and b, and H, b, and a, respectively.

Since sin(A) = h/c, h = c * sin(A).

Since cos(A) = b/c, b = c * cos(A).

The area of the triangle can be expressed in two ways:

One half the product of the base and height: (1/2) * b * a
One half the product of the sides: (1/2) * c * h
Substituting b and h in terms of c and A from steps 3 and 4 into the area formula and simplifying gives:
(1/2) * c * a * cos(A) = (1/2) * c * b * sin(A)
Simplifying further gives:
a * cos(A) = b * sin(A)
Squaring both sides of this equation and using the identity sin^2(A) + cos^2(A) = 1, we get:
a^2 * cos^2(A) + b^2 * sin^2(A) = b^2 - b^2 * cos^2(A)
Simplifying further gives:
a^2 * cos^2(A) + b^2 * sin^2(A) = b^2 * sin^2(A)
Dividing both sides by sin^2(A) and using the identity sin^2(A) + cos^2(A) = 1, we get:
a^2 / sin^2(A) + b^2 / cos^2(A) = b^2 / sin^2(A)
Multiplying both sides by sin^2(A) and simplifying gives:
a^2 + b^2 * cos^2(A) = b^2
Using the identity cos^2(A) = 1 - sin^2(A), we get:
a^2 + b^2 * (1 - sin^2(A)) = b^2
Simplifying further gives:
a^2 = b^2 * sin^2(A)
Dividing both sides by sin^2(A) gives:
a^2 / sin^2(A) = b^2
Using the identity sin^2(A) + cos^2(A) = 1, we get:
a^2 / (1 - cos^2(A)) = b^2
Simplifying further gives:
a^2 / sin^2(A) = b^2 / cos^2(A)
Substituting b / cos(A) for a * sin(A) from step 7 into this equation and simplifying gives:
b^2 = a^2 + b^2 - 2 * a * b * sin(A)
Rearranging terms gives:
a^2 + b^2 = c^2
This is the Pythagorean Theorem, which we have proved from trigonometry without circular logic.

Therefore, we have shown that the Pythagorean Theorem can be derived from trigonometry without resorting to circular logic.

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