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I have a small question regarding proposition 37 of the elements of Euclid. http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI37.html
The only problem I got with the proof is the fact that we don't seem to prove that we do have parallelogram. We have a figure with 4 side and we know that each opposite sides are parallels to each other. Is this considered enough to acknowledge that we have parallelograms ?
Thank you!
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Yes. If you draw one diagonal to divide the quadrilateral into two triangles, it is not too hard to show these are congruent (side and two angles). Hence opposite sides are equal in length.
Bob
I've had a quick look at your other post but it will take a while to take in all the relevant propositions. I'll reply when I've had time to do this.
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Yes. If you draw one diagonal to divide the quadrilateral into two triangles, it is not too hard to show these are congruent (side and two angles). Hence opposite sides are equal in length.
Bob
I've had a quick look at your other post but it will take a while to take in all the relevant propositions. I'll reply when I've had time to do this.
So, we do have a Rhomboid ( a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled.) ? And my question was, why does Euclid think it's enough to just show that we have opposite parallels ? I mean, a square also have its opposite sides parallels and so does a rectangle. Shouldn't he show that we have also equal opposite side and equal opposite angles ? (The side must not be equilateral and not right angled)
Last edited by Al-Allo (2015-05-21 04:52:36)
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I don't know my Euclidean propositions as well as you, but I suggest you look back to prop. 34.
There are a number of ways of defining a parallelogram:
All definitions require 4 sides but then:
opposite sides equal in length
OR
opposite sides parallel
OR
one pair of opposite sides equal and parallel
OR EVEN
diagonals bisect each other
I expect there are more. You can quickly show these are equivalent by congruent triangles.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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