Setup: - Name the quadrilateral ABCD - Name the mid points: E, F, G, H Proof: 1. EF is midliner of triangle ABC ==> EF is parallel to and equal to 1/2 AC 2. HG is midliner of triangle ADC ==> HG is parallel to and equal to 1/2 AC (1) + (2) ==> EF // HG and EF = HG. So EFGH is Parallelogram.
Another short proof from geometry: Draw a diagonal dividing the quad into two triangles with a common base. Join the midpoints of the sides of a triangle and you form a line segment that is half the length of the base and parallel to it. Both of the constructed triangles then will give lines parallel to that common base and half its length. Parallelogram is proved.
@@DeJay7 You've to admit Thales proof. So, he needs to add "(according to Thales theorem)", then it's a valid proof. Also, it uses Euclide axioms consequent (transitivity of parallels), but those rarely appears in proof (accepted as is).
Beautiful proof, i thought it could simply be proven that it forms a parallelogram because of the midpoint theorem. The length of the line that connects the midpoints of the sides must always be half the length from the beginning to the end. And because its the same thing for the other set of sides. They are equal to half the line from the beginning to the end and are both parallel to it A//B, A//C => B//C 1/2 A = B, 1/2 A= C. B=C Thus, it must be a parallelogram.
How would you prove the parallelism of the other two sides though? Explicitly, a parallelogram has 2 pairs of parallel sides. All we've proven here is that this shape has two sides that are parallel. This doesn't prove its a parallelogram, for a trapezium has the same properties, right? I'm sure similar steps could be used to find this out, but I feel like this could have been worth mentioning! For next time perhaps.
Ah! I now realise where my misunderstanding came from. Not only did Mr. Woo prove their parallelism, but also their equal magnitudes, proving its nature as a parallelogram. I'll listen better, next time!
You can use this method to test the other two sides as well. Just use two different vectors that are also equal (i.e. a-c and b-d) and halve them both. You'll find that the halved vectors are also parallel and in fact correspond to the other two sides.
hi eddie. i am a maths teacher. i see your videos and adopt the identical methods to teach my students. is it plaigarism?? will you sue me for copyright violation?
It's not a simple addition. It is an addition as vectors. a, b, c, and d are all vectors that enclose a quadrilateral, in that order. To get from the start of a to the end of b, it is the same net displacement as it is to follow the path along c, and then along d. That's how we have equal vector sums along a and b, as we do along c and d.
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I would have given a standing ovation after each class I attended. 🙏 You're fabulous.
Great teaching style. Clear and easy to follow.
At one point the answer came to me and I wanted to start shouting out the answer.
I love your teaching now I am also a teacher thank you sir
What an elegant proof! Just beautiful.
Amazing proof out of the blue. Just 2 lines, fabulous.
sir , you need to stop teaching students and start teaching teachers all over the world on ' how to teach '-❤️
2:32 So hogwarts wasnt enough education for Hermione
lmaoo 😂
Amazing proof! Thank you very much!
Fantastic !!!! Great explanation and easy to follow.
Sir your a magician ! ❤️
My mentor
Setup:
- Name the quadrilateral ABCD
- Name the mid points: E, F, G, H
Proof:
1. EF is midliner of triangle ABC ==> EF is parallel to and equal to 1/2 AC
2. HG is midliner of triangle ADC ==> HG is parallel to and equal to 1/2 AC
(1) + (2) ==> EF // HG and EF = HG. So EFGH is Parallelogram.
Another short proof from geometry: Draw a diagonal dividing the quad into two triangles with a common base. Join the midpoints of the sides of a triangle and you form a line segment that is half the length of the base and parallel to it. Both of the constructed triangles then will give lines parallel to that common base and half its length. Parallelogram is proved.
My mind was blown by how simply you put it and it seemed to make sense. I wonder if it could be proof.
@@DeJay7 You've to admit Thales proof. So, he needs to add "(according to Thales theorem)", then it's a valid proof.
Also, it uses Euclide axioms consequent (transitivity of parallels), but those rarely appears in proof (accepted as is).
Beautiful proof, i thought it could simply be proven that it forms a parallelogram because of the midpoint theorem.
The length of the line that connects the midpoints of the sides must always be half the length from the beginning to the end.
And because its the same thing for the other set of sides.
They are equal to half the line from the beginning to the end and are both parallel to it
A//B, A//C => B//C
1/2 A = B, 1/2 A= C. B=C
Thus, it must be a parallelogram.
Love it
Good teaching
Super explanation sir
hi eddieeeeeeeeee ur very helpful
Midline properties of triangle is an easier proof. But it is nice to see a vector proof.
You then need to prove midline of triangle or write you are using Thales theorem.
How would you prove the parallelism of the other two sides though? Explicitly, a parallelogram has 2 pairs of parallel sides. All we've proven here is that this shape has two sides that are parallel. This doesn't prove its a parallelogram, for a trapezium has the same properties, right?
I'm sure similar steps could be used to find this out, but I feel like this could have been worth mentioning! For next time perhaps.
Ah! I now realise where my misunderstanding came from. Not only did Mr. Woo prove their parallelism, but also their equal magnitudes, proving its nature as a parallelogram. I'll listen better, next time!
i wonder if people subscribe just to dislike the video, like, it's just 10 minutes and already a dislike
Yeah, there will always be some people trying to spread negativity. I feel like it might be a misclick though
Either misclick or disliking to remove these videos from their feed
But how do you KNOWWWWRRR it’s a parallelogram?
You can use this method to test the other two sides as well. Just use two different vectors that are also equal (i.e. a-c and b-d) and halve them both. You'll find that the halved vectors are also parallel and in fact correspond to the other two sides.
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hi eddie. i am a maths teacher. i see your videos and adopt the identical methods to teach my students. is it plaigarism?? will you sue me for copyright violation?
what kind of stupid ass question is this
Sir but how a+b=c+d ?
It's not a simple addition. It is an addition as vectors. a, b, c, and d are all vectors that enclose a quadrilateral, in that order. To get from the start of a to the end of b, it is the same net displacement as it is to follow the path along c, and then along d. That's how we have equal vector sums along a and b, as we do along c and d.
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