Thank you for all your help. Really helped through my A level maths and further maths but I wanted to ask if there were gonna be more videos on 1st and 2nd order DE's later on? Once again thank you ❤❤
丂LAYER, probably you are not the only one who wants to see a video on 1st and 2nd order DE's. Would you like to start a petition on Ablebees requesting Eddie to make a video on this topic? Others would support your petition.
The geometric proof needn't be that complicated - if you join O to P it's trivial to show the smaller triangles at the top and bottom are similar to the two larger triangles created by OP, and because those smaller triangles are formed by joining the midpoints, the sides similar to OP must each be half the length of OP and parallel to OP (and therefore each other). But yes, using vectors gets you there faster since vectors incorporate both direction and magnitude.
was this an online class or something which he recorded???cause i realy like the old ones which he recorded in the classrooms...not that i dont like these'
Can't I just split the quadrilateral horizontally and vertically and show that I have two pairs of parallel mid segment which is the definition of a parallelogram!
2:09 How is it that in the 21st century a mathematics teacher with an iPad does not have a software tool for measuring and/or marking the midpoint of a line segment? Using tilde to denote a vector is new to me but it seems to be becoming common since it's listed as a use for tilde and examiners are deducting points for not using it. I'm wondering if the proof would be any shorter using Geometric Algebra notation.
If you want to be rigorous then as others have commented, you simply switch the choice of origin and end point and equate two separate vector sums in the same way as was done here. But we don't really need to do that. We are seeking to prove that the midpoints join up to form a parallelogram, and that is proved once we show that *either* pair of opposite sides are both parallel and equal in magnitude. Once you show that, you've implicitly met the definition of a parallelogram, since if one pair of opposite sides are both parallel and equal in magnitude, the other pair must be also. Try it for yourself, if that condition is met, it's impossible to create any other type of quadrilateral. Of course the shape might be a rhombus or square, but these are merely special cases of a parallelogram.
