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CTK Exchange
mar
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Feb-18-07, 01:59 PM (EST) |
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"geometry"
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IS THERE ANY FOUR-SIDE SHARE WITH EQUAL THE OPPOSSITE ANGLES AND SIDES BUT IT'S NOT PARALLELOGRAMME;IF YES,COULD YOU DESCRIBE IT TO ME9PROBABLY YES).I HAVE MADE MANY THOUGHTS AND THIS QOESTINO HAS PROBABLY REFERENCE TO THEBAULT'S LEMME(NOT THEOREM) |
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alexb
Charter Member
1966 posts |
Feb-18-07, 02:08 PM (EST) |
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1. "RE: geometry"
In response to message #0
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>IS THERE ANY FOUR-SIDE SHARE WITH EQUAL THE OPPOSSITE ANGLES Fact 1: a quadrilateral with two pairs of equal opposite angles is a parallelogram. (Because then the opposite sides are parallel.) >AND SIDES Fact 2: a quadrilateral with two pairs of equal opposite sides is a parallelogram. (Because of SSS when you draw one of the diagonals.) There is a quadrilateral with two pairs of equal sides: a kite. It does have a pair of opposite equal angles. BUT IT'S NOT PARALLELOGRAMME; Not every kite is a parallelogram. |
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mar
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Feb-18-07, 07:24 PM (EST) |
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3. "RE: geometry"
In response to message #0
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>IS THERE ANY FOUR-SIDE SHApE WITH EQUAL only one pair of THE OPPOSSITE ANGLES >AND only one pair of the oppossite SIDES BUT IT'S NOT PARALLELOGRAMME;(i mean that i want an anti-example to the suggestion:if a four-side shape has only one pair of equal the oppossite sit's and only one pair of equal the oppossite angles,then it's parallelogramme)IF YES,COULD YOU >DESCRIBE IT TO ME PROBABLY YES).I HAVE MADE MANY THOUGHTS >AND THIS QuESTIon HAS PROBABLY REFERENCE TO THEBAULT'S >LEMME(NOT THEOREM)
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mar
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Feb-24-07, 11:50 PM (EST) |
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8. "RE: geometry"
In response to message #4
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I think that's the solution demanded.However,ADC'C is rectangular,isn't it-AD and CC' are parallel and in case they "meet" in point K,after comparing the triangles ADK and KC'C they are equal (AAA) so C'C and AD are equal-.How that's possible? |
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sfwc
Member since Jun-19-03
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Feb-25-07, 08:17 AM (EST) |
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9. "RE: geometry"
In response to message #8
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>I think that's the solution demanded.However,ADC'C is >rectangular,isn't it? No. See this page for a diagram>AD and CC' are parallel and in case >they "meet" in point K,after comparing the triangles ADK and >KC'C they are equal (AAA) so C'C and AD are equal-.How >that's possible? ADK and KC'C are similar (by AAA), but not congruent. So we cannot deduce that C'C = AD. Thankyou sfwc <>< |
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Kenneth Ramsey
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Feb-20-07, 09:49 PM (EST) |
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6. "RE: geometry"
In response to message #5
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>If the question is about quadrilaterals and not polygons, >meaning that the side intersections are allowed, then, for >an isosceles trapezoid ABCD with AB = CD, the quadrilateral >ABDC has a pair of equal opposite sides and two pairs of >equal opposite angles. My understanding is that an isosceles trapezoid by definition is a two dimensional figure and a polygon because it has more than three sides. Also, this might be the counter-example that Mar was looking for, since it has two opposite sides that are equal and two pairs of opposite angles that are equal. However, I think that a three dimensional figure is a better counterexample since it can have two pairs of equal opposite sides and two pairs of equal opposite angles. I did not see the word polygon in the question. |
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alexb
Charter Member
1966 posts |
Feb-20-07, 10:05 PM (EST) |
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7. "RE: geometry"
In response to message #6
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>>If the question is about quadrilaterals and not polygons, >>meaning that the side intersections are allowed, then, for >>an isosceles trapezoid ABCD with AB = CD, the quadrilateral >>ABDC has a pair of equal opposite sides and two pairs of >>equal opposite angles. >My understanding is that an isosceles trapezoid by >definition is a two dimensional figure and a polygon because >it has more than three sides. A triangle is a polygon too. But I should not have brought up the polygons, especially because I meant to say "a simple quadrilateral", which I an not sure is the definition of a 4-polygon. >Also, this might be the >counter-example that Mar was looking for, since it has two >opposite sides that are equal and two pairs of opposite >angles that are equal. I believe, he looked for a single pair of each. >However, I think that a three dimensional figure is a better >counterexample since it can have two pairs of equal opposite >sides and two pairs of equal opposite angles. A matter of taste. > I did not see the word polygon in the question. Right. But somehow I had an image in my mind. I think we now have a complete solution: one for simple and the other for self-intersecting quadrilaterals.
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mpdlc
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Feb-25-07, 07:26 PM (EST) |
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10. "RE: geometry"
In response to message #0
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Being an old fashioned engineer the solution I describe below it is not mathematical demonstration but it will give you the procedure to draw our elusive quadrilateral in any case. First I consider our potential quadrilateral like a four bar mechanism with two bars of equal length, I called them R1 and R2 . The two other bars, without loosing generalization one I can draw horizontal I named it L, and the other remaining bar joining the two other edges of bars R1 and R2 , I called it B. I called p the angle between R1 and L , and I called q the angle between R2 and L, both angles measured with a counterclockwise criteria , and I made q > p. I called w the angle formed by B and L, it is obvious w equal to q - p Now we take the horizontal and vertical projections of the polygonal represented by the four bars mechanism it will render the two following equations:
B cos (w) = R2 cos(q) - R1 cos (p) + L (1) B sin (w) = R2 sin (q) - R1 sin (p) (2) Since R1 equal R2 we can consider both as unit length then L and B will represent the ratio L/R and B/R By division of the above equations and sub sequential ordering we get cot (w) = (cos(q) - cos (p) + L) / (sin (q) - sin (p)) Remembering w = q-p now we can write L = cos (p) - cos (q) + (sin (p) - sin (q)) cot (q-p) So with this handy formula let us say we want to draw a quadrilateral two opposing side equal to 100 mm and the opposing angles forming p = 30 degrees and for example q = 45 degrees that will give us a value for L = 93.185 mm. and value for B amounts 80.02 mm
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