Pythagorean theorem: Euclid’s proof reimagined

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This is a short, animated visual proof of the Pythagorean theorem (the right triangle theorem) following essentially Euclid's proof. This theorem states the square of the hypotenuse of a right triangle is equal to the sum of squares of the two other side lengths.

For a longer version of this video with a bonus dissection proof based on this proof, see

You can also see other Pythagorean theorem proofs here:

or check out my Pythagorean theorem playlist:

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#math #pythagoreantheorem #pythagorean #triangle #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #mathshorts #mathvideo

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Proof by _shear_ force of will
An unexpected _turn_ of events

narfharder
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Isn't it more beautiful to draw 3 more triangles that are congruent to ABC ∆ around the c² square?

the-boy-who-lived
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This is also why the Pythagorean Theorem isn't true in non-Euclidean Geometries, because in those sheers and rotations are not area preserving.

jonathanaarhus
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I kinda prefer the the regular proofs, they just feel more certain in my mind. You know c^2=(a+b)^2-2ab

charles
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I like your animation. I have a similar proof, using Geogebra for the two shifts and one rotation. Cavalieri Principle is involved in the shifts.

theoremus
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The Pythagorean Theorem is not restricted only to squares. It'll work for circled or any other polygon.

abryg
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I think this is more of an illustration than a proof as such

morkris
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As a math student & aficionado I love visual proofs. As a math teacher I’m not as big a fan because visual proofs are not as precise or rigorous. We may not see tiny errors in a visual proof, but a logical, algebraic proof deals with fundamentally exact relationships. For example, there are great approximations for the length of an ellipse, but it can’t be given exactly. This means visual “proof” can mislead us into accepting an approximation as an equivalence. Therefore, visual proofs need to be cross-checked with more rigorous methods.

STEAMerBear
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I prefer the shear up, translate, shear down proof. It's much easier to visualize and there's no overlap.

smylesg
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Here’s another one:
So, you can take 4 of the same triangle and arrange them so the outer edges form a square. This is 2ab. The middle part (negative space) is c^2. So, 2ab+c^2=(a+b)^2. So, c^2=(a+b)^2-2ab=a^2+2ab+b^2-2ab=a^2+b^2, hence a^2+b^2=c^2.

thepianokid
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Wait this is literally just Euclid's first theorem, which I'm studying in school now.

In case people didn't know, Euclid's first theorem states that the area of the square of a leg is equal to the area of a rectangle with 1 side being the hypotenuse and the other side being the projection of that same side on the hypotenuse

kirahen
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Amazing! Can this technique be used to prove the law of cosines, that
c^2 = a^2 + b^2 -2ab*cos(γ)?

I guess the rotation is where the cosine might come from!

_P_a_o_l_o_
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What program do you use to make animations?

Поэзия-лъ
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Are you animating using Manim? Can you please tell me how you render your videos in vertical aspect ratio please?

nuralddd
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wait a second...
inv: inverse
sincos: Product of Sin and Cos (if both functions have the same inputs)


simplified to
a²=ac•sin(arctan(a/b) + a²sincos(arctan(a/b)
a=c•sin(arctan(a/b) + a sincos(arctan(a/b)
a(1-sincos(arctan a/b) = c sin(arctan a/b)

Value of a would be (if arctan a/b would be given or changed into Ω)
= c sin(Ω)/(1-sincos(Ω)

I don't know what prompted me to check something

vennstudios
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The horse on the hypotenuse is equal to the sum of the horses on the other sides.

Very.Crazy.Math.Pistols
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"Suppose we have a right triangle, " that is where the problem begins. How do you know it is a right triangle? well, it fits the theorem; but wait, how do you prove the theorem? well, suppose we have a right triangle. You see the problem?
the Correct way to express the theorem should be like this: if you find a triangle that has a relationship expressed in the theorem among its three sides in length, it must be a right triangle.

PCHUANG-ykpw
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That's not a proof. There is no proof this edit is accurately moving the areas of the squares.

ImGandalf
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Does it work on scalene and isosceles?

SecndCe
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666 likes

I won't mess with it 😎

matthewtallent