Can someone help me with my proof in Geometry homework?
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No picture to be found.  \n\nAre M, N, O, P all on the same line?  \n\nIf so, NOQ is isoceles, which makes MNQ and QOP congruent angles.  Since two sides and the included angle uniquely define the third side, so MQ and QP are congruent. and MQP is isoceles.
703288
Cool geometry im learning this in class 2bad i cant help u cuz im faling -.-
863581
You have a couple of typos, I think.  As for the "proof", dividing by x - y is dividing by zero.  That is a no-no.  (In other words, a * 0 = b * 0 does not imply a = b.)
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You can give geometry a logical structure and build on the simplest of facts.\n\nThese basic facts are accepted as being true and they are called GIVENS.  Euclid, the Greek mathematician called them AXIOMS.\n\nSo,  in geometry we start with basic facts or givens such  as  'vertically opposite angles are equal.' \n\nFrom this small collection of givens you can begin to prove other facts in geometry.   For example, you can proove that the angles of a triangle add up to 180 degrees.\n\nYou can use these proofs to prove other more complicated 'facts' about geometric shapes etc.  The great thing is that when you try to do this you don't have to keep going back and prove all the other things.  This approach is called DEDUCTIVE PROOF.  \n\nIn mathematics proofs are considered absolute.  Once a proof has been made it will always be true.\n\nThe abbreviations and symbols that are used are essentially the same as algebra.  This is a key point because rather than dealing with numbers you use symbols and then the proof is true for all values.\n\nA proof has to be set out logically and ALL the steps should be shown.  As you build the proof you use symbols such the three triangular dots for 'therefore' and ==> for 'which gives.'  Words such as 'necessary'; 'sufficient' and 'converse' are also used.\n\nAt the end of the proof mathematicians used to write QED which is an abbreviation for the latin 'quod erat demonstrandum' which means 'which was to be shown or proved.'  I think that has fallen out of fashion and people write 'which was to be proved' at the end of end with the last line of the proof.\n\nThere are other methods for making a proof.
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Geometric -- multiplying the previous term by 4
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let a = b  \nMultiply both sides by a \na² = ab  \n Add (a² - 2ab) to both sides \na² + a² - 2ab = ab + a² - 2ab  \n Factor the left, and collect like terms on the right \n2(a² - ab) = a² - ab  \n Divide both sides by (a² - ab)\n2 = 1\n(The place where the proof goes wrong is the very last step.  If a² - ab = 0, then you would be dividing by zero, which is undefined.)
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You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r
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What your proof will consist of depends on which statements you are allowed to use as axioms. For instance, Euclid, in his Elements, _defined_ perpendicular lines as those that created congruent, adjacent angles, and thus a proof of this statement in his system would be superflorous.\n\nSome axiomatic systems in high school give you the twin statements "the angle between perpendicular lines is 90°" and "a straight angle is 180°". In which case your proof would be:\n\nLet a and b be perpendicular lines, and P be their point of intersection. Let A be a point other than P on a, and B and C be on b such that P is between B and C. Then APB and APC are adjacent angles (as they share a common side, and niether one is within the other) which therefore sum to angle BPC. Since BPC is a straight angle, it is 180°. Further, since a and b are perpendicular, at least one of APB and APC are 90°, so the other must be 180°-90°, which is 90° also. Since both angles are 90°, they are congruent. Q.E.D.
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You can try these websites:\nwww.math.com/homeworkhelp/Geometry.html \nwww.mathleague.com/help/geometry/geometry.htm
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A regular n-gon has the centers of its inscribed and circumscribed circles the same point. If you draw segments from this center to each of the vertex, you will get n number of radii and n number of triangles, which are all congruent. Using derivation you will arrive at the conclusion that\nA = .5aP or A = .5ans where\nA is the area, a is the apothem, s is the side and P is the perimeter\na =2A/ns\nThe radius is acquired by\nsin [180degrees(n - 2)/n] = a/r\nr = a/sin[180degrees(n - 2)/n]
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Unfortunately, the people above have misused geometry.\n\nThe way that you'd solve this in a geometry class would involved quite a bit of trigonometry. Here's a way to solve it without using as much. (I worked it out both ways and both methods give the same result. I also checked the result.)\n\nI'm assuming that your octagon is regular.\n\nDraw a regular octagon. Call the length of each side a.\n\nNow use vertical and horizontal lines to draw gridlines on the octogon. This divides the octagon into four triangles, four rectangles, and a square in the middle.\n\nThe square has area a^2..\n\nThe triangles are 45-45-90 triangles having hypotenuse a. This means that both sides are a / square root of 2. This means that each triangle has area:\n\n1/2 * (a / sqr(2)) * (a / sqr(2)) = a^2 / 4.\n\nThe rectangles have one side that is a side of the octagon (a) and one side that is a leg of the triangle (a / sqr(2)).\n\nThis means that the rectangles have area:\n\na * (a / sqr(2)) = a^2 / sqr(2).\n\nThe octagon's area is equal to the sum of the areas of its regions:\n\n4 triangles + 4 rectangles + 1 square\n\n4*(a^2 / 4) + 4*(a^2 / sqr(2)) + a^2 = \n\na^2 + 2sqr(2)*a^2 + a^2 = \n\na^2 (2 + 2sqr(2)) = 64.\n\nSolve for a, and you get approximately 3.6407 feet.\n\nThat is to say, an octagon having a side length of 3.6407 feet has the same area as a square having a side length of 8 feet.\n\nThis seems contradictory until you realize that an regular octagon isn't made up of just squares and triangles. If they were, then the "diagonal" sides would be longer than the "horizontal" and "vertical" sides.
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First of all, u do not need Lagrange's identity to prove it. u just need to expand everything.\n\nSo, u want to use Lagrange's identity, fine. set a^m=a^2, a^n=x^2, substitute these into the Lagrange's identity and u get exactly [a^{(m+n)/2}+b*y]^(2) + [y*{a^(m/2)} - {a^(n/2)}*b]^(2) without even needing to do anything else.\n\nThis is not called proving!
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Try this link and read it all the way through.  It's short.\n\nhttp://mathcentral.uregina.ca/qq/database/QQ.09.00/tom3.html\n\nThis link is better:\n\nhttp://mathforum.org/library/drmath/view/56760.html\n\nJust scroll down.
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There are quite a few method for proving pythagoras theorem. I feel that easiest and neatest is by similarity of triagles ..\n\nproof 6 of the link
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http://www.studentquestions.com/homework_ans/4244.html
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There is no work to show, rather theorems to justify the solutions.\n\nInscribed angles of a circle that subtend a diameter are 90 degrees. This is a corollary of a theorem which states that inscribed angles are 1/2 the measure of a sector angle which subtends the same arc. In the diameter case the sector angle is 180 degrees, thus the inscribed angles are 90.\n\nThe quadrilateral that contains angle Z has it's vertices on the circumference and is cyclic. A cyclic quadrilateral has the property that opposite angles are supplimentary (add  up to 180) thus angle Z = 180 - 148 = 32 degrees.\n\nI am not sure how you would show work toher than proving the theorems.\n\nHope it helps
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Why don't you share with us what you have done so far instead of just asking someone to do your homework?\n\nSorry about the snide remark. Here is your well-deserved step-by-step solution:\nhttp://www.rgetso.com/yahoo/afchica101.pdf