I don't remember ever learning this proof. Then again, it was more than 20 years ago... Thank you! It was so satisfying to see how it all comes together for the last part!
You'll find it easier to explain geometry if you label significant points. Take the small arc subtending the angle θ at the centre and mark its midpoint C. Join that point to each of the ends of the small arc (A and B). The quadrilateral formed by A, B, C and the centre (O) is a kite, since AC=AB and OA=OB, radii. Now join OC, which bisects θ, and is another radius. Now you have two congruent isosceles triangles, OAC and OBC, so angle OAC=OCA=(180°-θ/2)/2=90°-θ/4, and similarly OBC=OCB=90°-θ/4. Therefore angle ACB = ACO + BCO = 180°-θ/2. Now pick any point D on the large arc from A to B. Join AD and BD. Then ACBD is a cyclic quadrilateral whose opposite angles sum to 180°. So angle ADB = 180°-ACB = 180°-(180°-θ/2) = θ/2, proving the theorem.
Case 1: θ is the external angle and the two opposite angles are α (isoceles, as you said), so you get θ = α + α = 2α directly, since the external angle is the sum of the two opposite angles. I think it more or less boils down to the same thing, or am I barking up the wrong tree? That's the trouble I have with axiomatic systems: it's never quite clear which ones you are allowed to assume. If that was a bit opaque, look up the Feynman lecture where he talks about it. Most likely I misunderstood that, too ......
There is an easy way using isosceles triangles. No cases needed. The diameter is just a special case when the sector angle 2(x + y) = π which forces the inscribed angle to be right.
I don't think I learned this in school; but then I might have forgotten about it. There are just so many theorem concerning triangles! I did learn, and do remember!, Thales's theorem, though, which is a special case of this theorem as I just realized (the central angle being 180° degrees -a straight line: the diameter of the circle- and the inscribed angle being 90° degrees)! 😀
You can view this as a generalization of the theorem that says that if you have a triangle with a base that is a diagonal of a circle and another vertex that is on the circle the angle at that vertex is 90 degrees. Replace that diagonal with an arbitrary chord and then the angle at the other vertex is no longer 90 degrees but it is a constant for a given chord. The version with the diagonal is trivial to prove using the Pythagorean theorem. When I first saw this I tried to prove it geometrically and utterly failed. Then I tried to use analytic geometry, and come up with a complex and ridiculous formula that should be zero iff the theorem is true. Mathematica said that the formula was indeed 0, but I could not manage to prove that, so again utter failure. Then I tried an approach based on vectors, which led to yet another ridiculously complex formula that should be zero, which again I could not prove but Mathematica said it was zero. Then I tried turning the theorem around. Instead of fixing a chord and moving the other vertex around, I imagined fixing the other vertex and sliding the chord around. The theorem is saying that the angle at the vertex is constant when you do that. So I started visualizing the circle as a giant circular train track, with the train going around the circle at a constant angular velocity. If you were at the center of the circle and had two cameras that you wanted to have track the front and back of the train, respectively, as it went around the circle you could do that by putting both cameras on a single turntable, and rotate that circle at the same angular velocity as the train. If you were standing outside the circle, or inside the circle but not on the center, you would need separate turntables because you would not see the same angular velocity at the two ends of the train, and the turntables would have to be variable speed. But if you were on the circle what the inscribed angle theorem is saying that you would only need one turntable, and it would only need constant angular velocity (1/2 the angular velocity that an observer at the center sees). This has to work for any pair of points on the train, not just the front and back, and so that means we can get rid of the train and just consider a point traveling at constant speed around the circle. The theorem is saying that for such a point an observer on the circle will see the point moving at a constant angular velocity that is 1/2 the angular velocity that an observer at the center of the circle would see. Put in that form it is easy to set up the equations of motion for a particle in a circular orbit around a point, then figure out the equations for its motion in polar coordinates centered at a point on the circle, get the angular component of the velocity, and that turns out to actually be something simple enough to easily see it is constant. So finally I actually had a proof of the damn thing, although I had to turn it into a physics problem to do it.
IMHO, there's another case to be made, where the point is between the other points on the arc! You can probably show, that in this case, (360° - θ) = 2 α. Proof by, who would have thouht, case 1.
If you have 3rd case for example, can you just take alpha and drag it around to make the 1st case? The angle stays the same, so that can also be another way to prove both 2nd and 3rd case using 1st case only?
@@cyrusyeung8096 Correct! It's easy to see, if you make it about the arc (360° - θ) instead of θ. Then the minor arc is the "outer" arc again, and cases 1 to 3 hold with (360° - θ) and 2 α being equal again!
You just use the fact that opposite angles of a cyclic quadrilateral sum to 180°. The two end-points of the arc subtending the angle θ at the centre, any point on the major arc, and any point on the minor arc make a cyclic quadrilateral. Call the angle subtended by the end-points on the minor arc β. So β = 180°-α = 180°-θ/2. The reflex of that angle is 360°-β = 360°-(180°-θ/2) = 180+θ/2 = 180°+α
@avinotion I am pretty good at math, I just know that half of math is show and tell, and the other half is theory and practice. I agree that anyone can just rewind a finite amount of times, but also teachers can just use mathematical programs to begin with. like geogebra. because making some random drawing is very old school. teaching people how to use these mathematical tools is part of teaching. and if there is pressure on student to do better there need to be pressure on teachers to be better.
You know what it was a good explanation but you could get a better in channel cuemath ( is unpopular but you would probably get shocked ) they generally post videos on proofs ( not disrespecting this channel because it's one of my favourite channel .
Please DO NOT allow UA-cam to interrupt you're carefully prepared videos with worthless and annoying pop-up ads. Just when one is really concentrating on your fantastic content POW!!! Something totally unrelated pops up on the screen. Result? The thinking process is interrupted at a critical juncture - NOT conducive to learning.
How to find the area of the circumcircle of a triangle: ua-cam.com/video/w0JFaEwYpNE/v-deo.html
I don't remember ever learning this proof. Then again, it was more than 20 years ago...
Thank you! It was so satisfying to see how it all comes together for the last part!
Case three is very interesting.
You'll find it easier to explain geometry if you label significant points. Take the small arc subtending the angle θ at the centre and mark its midpoint C. Join that point to each of the ends of the small arc (A and B). The quadrilateral formed by A, B, C and the centre (O) is a kite, since AC=AB and OA=OB, radii. Now join OC, which bisects θ, and is another radius. Now you have two congruent isosceles triangles, OAC and OBC, so angle OAC=OCA=(180°-θ/2)/2=90°-θ/4, and similarly OBC=OCB=90°-θ/4. Therefore angle ACB = ACO + BCO = 180°-θ/2. Now pick any point D on the large arc from A to B. Join AD and BD. Then ACBD is a cyclic quadrilateral whose opposite angles sum to 180°. So angle ADB = 180°-ACB = 180°-(180°-θ/2) = θ/2, proving the theorem.
Case 1: θ is the external angle and the two opposite angles are α (isoceles, as you said), so you get θ = α + α = 2α directly, since the external angle is the sum of the two opposite angles.
I think it more or less boils down to the same thing, or am I barking up the wrong tree? That's the trouble I have with axiomatic systems: it's never quite clear which ones you are allowed to assume. If that was a bit opaque, look up the Feynman lecture where he talks about it. Most likely I misunderstood that, too ......
There is an easy way using isosceles triangles. No cases needed. The diameter is just a special case when the sector angle 2(x + y) = π which forces the inscribed angle to be right.
Wow, thank you so much!
I don't think I learned this in school; but then I might have forgotten about it. There are just so many theorem concerning triangles!
I did learn, and do remember!, Thales's theorem, though, which is a special case of this theorem as I just realized (the central angle being 180° degrees -a straight line: the diameter of the circle- and the inscribed angle being 90° degrees)! 😀
You can view this as a generalization of the theorem that says that if you have a triangle with a base that is a diagonal of a circle and another vertex that is on the circle the angle at that vertex is 90 degrees. Replace that diagonal with an arbitrary chord and then the angle at the other vertex is no longer 90 degrees but it is a constant for a given chord. The version with the diagonal is trivial to prove using the Pythagorean theorem. When I first saw this I tried to prove it geometrically and utterly failed.
Then I tried to use analytic geometry, and come up with a complex and ridiculous formula that should be zero iff the theorem is true. Mathematica said that the formula was indeed 0, but I could not manage to prove that, so again utter failure.
Then I tried an approach based on vectors, which led to yet another ridiculously complex formula that should be zero, which again I could not prove but Mathematica said it was zero.
Then I tried turning the theorem around. Instead of fixing a chord and moving the other vertex around, I imagined fixing the other vertex and sliding the chord around. The theorem is saying that the angle at the vertex is constant when you do that. So I started visualizing the circle as a giant circular train track, with the train going around the circle at a constant angular velocity. If you were at the center of the circle and had two cameras that you wanted to have track the front and back of the train, respectively, as it went around the circle you could do that by putting both cameras on a single turntable, and rotate that circle at the same angular velocity as the train.
If you were standing outside the circle, or inside the circle but not on the center, you would need separate turntables because you would not see the same angular velocity at the two ends of the train, and the turntables would have to be variable speed.
But if you were on the circle what the inscribed angle theorem is saying that you would only need one turntable, and it would only need constant angular velocity (1/2 the angular velocity that an observer at the center sees).
This has to work for any pair of points on the train, not just the front and back, and so that means we can get rid of the train and just consider a point traveling at constant speed around the circle. The theorem is saying that for such a point an observer on the circle will see the point moving at a constant angular velocity that is 1/2 the angular velocity that an observer at the center of the circle would see.
Put in that form it is easy to set up the equations of motion for a particle in a circular orbit around a point, then figure out the equations for its motion in polar coordinates centered at a point on the circle, get the angular component of the velocity, and that turns out to actually be something simple enough to easily see it is constant. So finally I actually had a proof of the damn thing, although I had to turn it into a physics problem to do it.
Hey these could be also proved by exterior angle theorem of triangle that would be easier that letting that alpha or theta
IMHO, there's another case to be made, where the point is between the other points on the arc!
You can probably show, that in this case, (360° - θ) = 2 α. Proof by, who would have thouht, case 1.
You can also write it as θ = 2 (180° - α) = (360° - 2 α) or (θ + 2 α) = 360°.
If you have 3rd case for example, can you just take alpha and drag it around to make the 1st case? The angle stays the same, so that can also be another way to prove both 2nd and 3rd case using 1st case only?
it's a pretty intuitive thing in real life
like you get 2x closer to something and it appears 2x bigger
I don't remember this one. It would have been well over 50 years if I did learn it.
Nice, I like these types of videos
It can be shown by playing with isosceles triangles
What if the third point was infinitely close to one of the other two points? Wouldn't that make it approach either 0° or 180°?
Yes, my intuition says the angle must close as we approach one of the minor arc end points. But... the proof is proof!
2:01
What happened to you standing in front of a whiteboard? I want it back, seeing you helps in following along, I think.
What about a point inside the minor arc
The relationship still kind of holds but reflex angle is considered. That is, 360° - θ = 2α.
@ ahh yes thank you
@@cyrusyeung8096 Correct! It's easy to see, if you make it about the arc (360° - θ) instead of θ. Then the minor arc is the "outer" arc again, and cases 1 to 3 hold with (360° - θ) and 2 α being equal again!
You just use the fact that opposite angles of a cyclic quadrilateral sum to 180°. The two end-points of the arc subtending the angle θ at the centre, any point on the major arc, and any point on the minor arc make a cyclic quadrilateral. Call the angle subtended by the end-points on the minor arc β. So β = 180°-α = 180°-θ/2.
The reflex of that angle is 360°-β = 360°-(180°-θ/2) = 180+θ/2 = 180°+α
the third case was much more unclear. the first two was good and easy but it seems like the third one needs more precise show and tell.
I recommend replaying it a couple more times.
In short, case 2's θ1 and θ2 are now case 3's θ and θ1
@avinotion I am pretty good at math, I just know that half of math is show and tell, and the other half is theory and practice. I agree that anyone can just rewind a finite amount of times, but also teachers can just use mathematical programs to begin with. like geogebra. because making some random drawing is very old school. teaching people how to use these mathematical tools is part of teaching. and if there is pressure on student to do better there need to be pressure on teachers to be better.
New way of making video unlocked 🔓
Angle at center is twice angle at circumference. O level maths
Proving the proof
I just proved it in front of the class by myself last week
Did you have to do three cases, or were you shown an easier way?
@ Yep those 3 ways
I tried proving it using vectors and i did not cook
You know what it was a good explanation but you could get a better in channel cuemath ( is unpopular but you would probably get shocked ) they generally post videos on proofs ( not disrespecting this channel because it's one of my favourite channel .
Cool
Please DO NOT allow UA-cam to interrupt you're carefully prepared videos with worthless and annoying pop-up ads. Just when one is really concentrating on your fantastic content POW!!! Something totally unrelated pops up on the screen. Result? The thinking process is interrupted at a critical juncture - NOT conducive to learning.
It's not his fault. Only way to solve your problem is UA-cam premium.
@@ianfowler9340or ublock origin
It will probably get resolved when you get your homophones sorted out.
And u block origin
does this mean the host is making profit using this video?? but he deserves it😢