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Intersecting Chords Theorem (visual proof)
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- Опубликовано: 28 сен 2022
- This is a short, animated visual proof demonstrating and proving the intersecting chords theorem from geometry. #manim #math #mathshorts #mathvideo #intersectingchords #geometry #chords #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #inequality
This animation is based on a visual proof from Hubert Shutrick that appears on the beautiful math website Cut-the-knot (www.cut-the-knot.org/Curricul...)
To learn more about animating with manim, check out:
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This channel is seriously underrated. I've always loved showing visual proofs when tutoring math, as I think it illustrates how beautiful math can be and can make things a lot more intuitive and clear up a lot of confusion that comes with dense notation and jargon. Thanks for another great video.
Thanks for watching! I agree that it can't hurt to have a visual representation to go along with "formal" ideas :)
@@MathVisualProofs I agree that "Mayor McCheese is a fun name
Great proof, and a nice visualization.
Thanks as always! :)
Mind blown away. The idea of scaling and attaching the triangles together was fresh one to me. I would have never thought of that
:) Glad you liked it!
agreed, that was pretty sweet.
@@RandyKing314 thanks :)
I used to hate proofs. Your channel changed my mind, and that's saying a lot!
😀
Hey so I did some thinking: firstly - I stumbled on a proof that didn’t need scaling and just uses setting the ratio of sides of one triangle to ratio of sides of another to finish proof without ever needing scaling.
That being said - I am still uneasy and untrusting of the idea of “scaling” to use the proof you did. I can’t explain WHY it makes me uneasy but can you help me feel I can trust your scaling idea? Thank you and again - love you channel!
Thanks to you, I can understand it👍
Thank you!!!
Welcome!
@@MathVisualProofs Nice animations! I knew this theorem but just had to watch again because you made it so well;)
excellent!!!
Thank you! Appreciate it.
What a gorgeous video and great channel! Thank you so much for helping those with Aphantasia who want to grasp geometry!
Glad it helps!
@@MathVisualProofs would you please perhaps use a different approach to explain to me why I can trust the scaling method? I saw another method proving interesting chiefs theorem using ratio of sides of similar triangles set equal to one another and no scaling used. Please help me trust scaling as legitimate. I want to open my eyes to your truth.
@@MathCuriousity If you are using similarity of triangles, you are essentially using the scaling argument. The ratio of sides being equal is essentially saying you can scale one triangle to the other. In this visual argument, we use scaling so we can create the parallelogram. So we scale both triangles independently to similar triangles that then fit together along a diagonal.
@@MathVisualProofsI didn’t think of the similar triangles also z using scaling. Very interesting. Let me reapproach your solution with that and then get back to you. Thank u so much!❤
Amazing
Thanks!
Excellent
Thanks!
Thank you for the video.
Thanks for watching
How about a proof that uses Ptolemy's theorem?
Excellent ❤️👍
Thanks!
Hey question: firstly love your channel and love these proofs, however - the moment we scale a triangle up, aren’t we changing the triangles and thus providing a faulty proof? Could you help me understand why your scaling of both was “legal” and sort of preserved the integrity of the proof?
As long as you scale each side by the same value the triangles maintain their angles and that’s what’s important for the proof. The angles fit together so that we have a parallelogram.
And thanks!
@@MathVisualProofs I think I see. Thank you kindly!
Angle Side Side theorem says that if two triangles have the same corresponding angle and and same corresponding two sides. In this case this is true we have congruent triangles. Flipping the upper left triangle up and translating across a line ab coresponds to cd. Thats why ab = cd
Two trianles are simmilar. So ratios of b/d = c/a, multipying both sides by a*d, you get a * b = c * d
So could anyone explain? I draw a circle 10cm diametre. Then a line from circle to the center with 7 cm length. Now a line perpendicular the free extreme 4,5cm length. So I want to know what's the diameter.
axb=dxc, so 7b=4,5x4,5, 7b=20,25; b=20,25/7= 2,89cm.
But I know it's 3cm. 😮
Yoo who else gotta do this in class
While clearly proven true, I find this unsatisfying as a visual proof. First(and most importantly), it hinges entirely on the subtending angle theorem, which is not intuitive and should also be visually proven as a lemma, or at least reference to a separate visual-proof video for that fact. Second, the "scale each triangle by a length from the other" step isn't really visualized. Instead we just see them scaled to the same size arbitrarily(see below**). Third the parallelogram was unnecessary. It doesn't hurt, per se, but is a bit of a distraction. Simply the fact that the triangles are similar is enough to say all scaled sides are equal -- you might as well super-impose them. The way the parallelogram is presented makes it seem important/essential to the proof, which it isn't.
**On the arbitrary scaling: Depending on the size of the circle, scaling could shrink both, or one could become larger while the other becomes smaller. Just resizing two similar triangles to the same size doesn't really "visualize" the scaling by these factors, so the important fact that they rescale to the same size, is merely asserted, and not demonstrated visually. The assertion is visualized by an arbitrary choice of size, but the assertion itself, is not proven visually. A case could be made this step is so obvious it does not need to be visualized, but I'd say if you can't visualize the multiplication of lengths of two line segments, you need to at least mention that scaled sizes are unknown and show smoothly varying possibilities by putting actual length values, varying those, and showing the two scaled triangles always the same size from both being tiny, to one larger and one smaller, to both being larger.
Interesting thoughts. Thanks for sharing. I agree that this relies on another theorem, but I am fine stating that one and leaving people to look it up if they are interested (is how to learn best). As to the parallelogram argument, I like it that way - and I was staying true to the cited visual proof: www.cut-the-knot.org/Curriculum/Geometry/GeoGebra/HubertIntersectingChords.shtml :)
** "hinges entirely on the subtending angle theorem" ** I only ½ agree with this. For one pair of equal angles, vertically opposite angles can be used, and equality 2 angles of triangles imply their similarity.
@@ojas3464 Can be used, but wasn't. 😜
And you still _need_ the chord theorem.
I went to hit the like thumb, but noticed that I already had.
Too bad I can't do it again.
Hah! Thanks. I appreciate it regardless 👍😀
Now, show that all angles subtended by and arc must be equal.
top dollar..
:) 👍
May exam check
I had to watch twice to understand how you scale one triangle by b and the other by c. Explain it on more detail for the other viewers who may not understand as well as I did.
I think you mean cords ;)
not clear at all, the triangles all look different area
They aren't supposed to, the scaled triangles are