|WikiProject Mathematics||(Rated C-class, Mid-importance)|
My daughters has a teacher wants her to draw a Quadrangle that has 2 pairs of equal sides but is NOT a parallelogram. Can it be done? not at all. totally impossible —Preceding unsigned comment added by 184.108.40.206 (talk • contribs) 22:25, 18 September 2005
- See the figures at "quadrilateral": a kite a should do. (If you mean by quadrangle something which must have 4 angles but not necessaritly straigt lines as borders, there are many other possibilities.) — MFH:Talk 14:21, 24 March 2006 (UTC)
- An isosceles trapezoid can also fulfill the requirements. Opposing sides can be equal in length but only one facing side is parallel.
- I think you mean adjacent sides, and then your trapezium (trapezoid) turns into a kite. If you really meant opposite sides, then see the first characterisation in the article to see that you have a parallelogram. Dbfirs 15:35, 17 April 2017 (UTC)
Most definitions start off with the simple definition of a parallelogram as a quadrilateral with both facing sides parallel, yet this is not in the list of conditions that satisfy the definition of a parallelogram.
Doesnt sound right:
Proof that diagonals bisect each other
Wouldn't the congruent triangles proof be better as the main (and only) proof, rather than as alternative? Different schools of education seem to set out proofs in different ways. If I edit, I will probably upset schools in the USA! Dbfirs 08:55, 20 January 2009 (UTC)
- If no-one has a rational opinion on this, I'll just go ahead and edit. Dbfirs 12:21, 15 June 2009 (UTC)
- Cited the well known ASA postulate which is directly shown rather than an AAS condition. Revised the conclusion to match what was to be shown (ie QED). This is how I was taught to do proofs. The point about "E" (half-way along?) has nothing to do with what we are trying to prove, however I left it in by revising the wording to have it consistent with the preceding content. Discussion regarding "coordinate grid" is nonrelevant. I realize the latter content was probably left-overs from another proof.
- Thanks for your improvement (genuine thanks). The comment was not intended to be "smart-alek". Perhaps you didn't see what I was commenting on (a proof by similarity). I was genuinely leaving the setting-out of the proof for an American format because we do set out proofs differently here, though I'm glad you approve of most of the change I made, and thanks for improving the end. I'm sorry you are so sensitive about anti-American bigotry - there was no such intention here! We all tend to believe that what we ourselves were taught is the only correct way. I do tend to feel overwhelmed by America when Wikipedia reads like an American school text book and sounds so "foreign" to British readers, but this was not the case here. Dbfirs 00:01, 13 December 2009 (UTC)
While I can appreciate the idea of shortening things up, it doesn't work in this situation. The proof does not require nor use the concept of a midpoint for a line segment. In an actual 2-column proof, the last line would read something such as this, AC(vinculum) and BD(vinculum) bisect each other. Reason: If two segments divide each other into segments of equal length (US: congruent segments), they bisect each other. The midpoint is not part of the proof. I left the midpoint "add-on" in there because it was there previously. I now believe it should be removed since it is not part of this proof and may confuse the situation. Thoughts? JackOL31 (talk) 19:19, 14 February 2010 (UTC)
- Since the mid-point is specifically identified as part of the proof, I believe that it is clearer for the general reader to use this identified point, but if you think that the proof is "purer" without mentioning mid-point then I am happy to retain your wording. Dbfirs 22:23, 15 February 2010 (UTC)
Computing the area of a parallelogram
- Take the origin as one vertex, then (in two dimensions) (a1, a2) is an adjacent vertex, and (b1, b2) is the other vertex adjacent to the origin. The complete parallelogram is thus uniquely defined. In general, any two non-parallel vectors a and b uniquely define a parallelogram with long diagonal a + b and short diagonal a - b. Dbfirs 08:11, 10 August 2009 (UTC)
Let a,b be vectors , then the area of the parallelogram with adjacent sides represented by vectors a = (a1, a2, a3) and b = (b1, b2, b3) is given by absolute value of the vector product of a and b, i.e. by the determinant of a matrix built using the two vectors as columns, with the last column padded using ones as follows:
- This is then consistent with my 2-D explanation above. What do others think? Dbfirs 18:13, 10 August 2009 (UTC)
- ... (later) ... Actually, I've just realised that the original editor was thinking in terms of co-ordinate geometry, rather than vectors, so the letters represent co-ordinates of points: ... A = (a1, a2), B = (b1, b2) and C = (c1, c2), where A, B & C are any three vertices of any parallelogram in 2-D space. I'm more accustomed to thinking in terms of vectors, but perhaps this explanation is simpler for most people. It does need spelling out in the article for those not accustomed to mathematical notation. Dbfirs 18:33, 10 August 2009 (UTC)
I corrected the heading "Types of parallelograms" to read "Types of parallelogram". There's only one class of thing called "parallelogram", but with many types. If I broke your link, hey sorry.--Rfsmit (talk) 22:23, 25 February 2010 (UTC)
Properties before characterisations
The current order, with minimal properties in characterisations, is creating confusion. Don't most people expect to see a full list of properties first, before the "if and only if" conditions? On the other hand, we don't really want to repeat the list. What is the best way to show this clearly? Dbfirs 07:30, 18 December 2011 (UTC)