which postulate can be used to prove that triangle DEG &GFD congruent if DE &GF congruent, & EG & FD congruent
815401
There is no such postulate.
1085012
Its been a long time since I took Geometry but maybe I can give you a small pointer.\n\nTry dividing the rectangle into 3 segments lengthwise and 3 segments widthwise\n\nA--B--C--D\nE--F--G--H\nJ---K--L--M\nN--O--P--Q\n\nDraw the diagonal through NKGD\n\nProve that the segments NK, KG, and DG are equal (prove that the triangles are the same)\n\nNow draw the segments from the vertex Q to the mid point of AD and AN.  Prove that the mid point of AD is the same as the mid point of BC and the same for AN and EJ.\n\nNow we have to prove that the line segment we drew in the last step instersect points K and G. You can probably do this with similar triangles (the triangle NQ and mid point of EJ is similar to JK and mid point of EJ therefore the point K is in the triangle)\n\nSo if you prove that the lines drawn from Q to the midpoints intersect K and G and we proved that NK KG and GD are congruent we have proved the original hypothesis.\n\nYou can turn this gabably gook into a real mathematical proof.\n\nGood Luck
1130099
(√3)² + 1 ² = 2²\nsin α = √3/2 α = 60º;  ß = 90 - α;  ß = 30º\n\nAnd your second question is\naⁿ = c² - b².  I'm going to assume that you wanted to put in a, b and c are all integers and a > 1.  Because if not, your theorem is not true and cannot be proven.  With that stipulation, this theorem passes the "intuitive" test.  It makes sense.\nWe know for certain that, for all even n, it must be true by the Pythagorean theorem.  So only odd n are of consequence to this theorem.\n\nI'm going to take a moment here and say that if you're trying to use this theorem as a FLT lemma, you must understand that it won't fly.  The lemma must use the carry the same stipulations as the successor theorem.  Said the other way, the corollary is bound by the restrictions of the theorem.\n\na=2, n=3 gives us c=3 and b=1.  8=9-1.  I took some time trying to find a counter-example and could not.  I did find some neat patterns emerging.\n\nThe proof to your theorem is simple: aⁿ = a²[(½a+½)² - (½a-½)²] because the expression in brackets will always be equal to a (a is odd prime; composite a can be simplified to prime a).\n\nI also found that 2ⁿ= [2^(n-2)+1]² - [2^(n-2)-1]².  So, in addition to all the solutions from the above expression, there will be another set of solutions for 2ⁿ; n is odd integer > 5.
831013
suppose c is even.  then this means that a is even and b is even, or a is odd and b is odd.  that is, a and b have the same parity.  we could not have a even, b even and c even because otherwise, a, b and c would not be coprime.  so a is odd and b is odd, if c is even.  I am unable to do this right now (I have forgotten my math, sorry), but I believe you could dispense with a proof which, loosely speaking, allows you to conclude that c must be odd on the basis of the fact that c^n = 0 (mod 2) is too restrictive (i.e. not all nth powers are even (except for those containing 2^n as a factor), or in other words, most nth powers are odd).  sorry, am not much help.\n\nincidentally though (if I may mention it here), a proof for FLT has already been published (and verified) in 1994.  the original proof comprised a whole edition of the Annals of Mathematics for that year and is composed of two papers, to wit: \n\nModular Elliptic Curves and Fermat's Last Theorem\n\nAND\n\n(again, it is easy to forget article titles, so please just search on the WWW web for it... :-))\n\nthe proof is subtle and surprisingly involved, and there is actually a page on the Internet dedicated to presenting the mathematics behind the proof to the general mathematical audience.  try googling "The Mathematics of Fermat's Last Theorem".\n\nhappy reading!
153034
No, nobody really can prove it, with assuming several axioms\nfirst. There are certain things you have to assume are true in your universe, without necessary being able to check them. From those axioms, you can derive conclusions (theorems) that are always true, but only in that particular\nuniverse. The discussed operation of addition is one such theorem. If you assume that the axioms of natural numbers are true, then this addition becomes quite trivival. If you assume those axioms of natural numbers to be false, then the above operation of addition might become false with a different set of supposition. In other words, only with the facts given, one cannot prove, nor disprove, the fact of addition. It all depends on the set of assumptions you would have made before adding the two ones.
789629
While the above answerer is correct that there is an SAS theorem for proving triangles congruent, there is also an SAS theorem for proving triangles similar:  If two sides of one triangle are proportional to two corresponding sides of another triangle, and their included angles are congruent, then the triangles are similar.  So SAS can in fact be used to prove triangles are similar.
243852
Hope this helps\n\n\nhttp://argyll.epsb.ca/jreed/math9/strand3/triangle_congruent.htm\n\nhttp://mathforum.org/geopow/fullsolutions/102596.solutions.html or www.mathforum.org
665741
http://en.wikipedia.org/wiki/Pythagorean_theorem#Proofs\n\nCheck below for the "proof of converse".  It's the most straightforward there is.\n\nBasically, you try to construct a triangle with x,y, and z, only to discover that it is congruent to a right triangle!
222062
If u want to know if this triangle is possible or not,then the answer is Yes it is possible.\na triangle can be constructed if the sum of any two sides is always greater than the third side.\nand here   8+9>10\n           8+10>9\n           9+10>8
241805
its easy\nsince the triangles HOI and JOI are congruent\nwe will find when you will draw the figure on a paper\n\nangle HOI=angle JOI\nsince the triangles are congruent\n\nbut angle HOI+ angle JOI=180 degrees\n\ntherefore angle HOI & JOI=90 degrees\n\nand side H0=J0\n\nnow using the theorem we get since the diagonals bisect and intersect and make an angle of 90 degrees the parallelogram is a rhombus.\nnow for the 10 points
783069
Prove that if in an isosceles triangle a line segment does one of the following things it also does the other two.\n\n      A\n       ^\n    /  |  &#92;\n   -------\nB    D    C\n\nAD goes from the vertex to the midpoint of the base \nAD is perpendicular to BC \nAD bisects /A \n\n\nConverse: If the same line does any two of the previous three things, then the triangle is isosceles.
769341
Your first Question:\n\nNow,\n\nImagine a Triangle with vertices ABC. \nAny line from A to BC is AD. D is the point of intersection on BC.\n\nLet E and F be the midpoints of Sides AB anc AC respectively.\n\nJoin E and F to meet AD in G.\n\nNow for the proof:\nTake Triangles AGE and ADB these two are similar.\n\nReason being that EF would be parallel to BC from Mid Point theorem. Therefore, the two triangles will be similar as Angle E and Angle B are equal and Angle A is common. \n\nNow take Triangle ADB, E is the mid point of AB and EG is parallel to BD there for from Converse of Midpoint Theorem G becomes the Midpoint of AD.\n\nHence proved.
673009
Not all triangles are congruent.
740926
Use graph paper and graph out this parallelogram..\nA(-6,0) B(0,10) C(6,0) D(0,-10)\nAngle A = ang C (definition of parallelogram)\nAngle B = ang D (def. of para)\nThe diagonals bisect the angles(given)\nThe diagonals bisect each other(def. of Para)\nUsing side-angle-side you can prove all the triangles formed by the diagonals are congruent. So that means all sides of parallelogram are equal and you have a rhombus.
447231
Aha, geometry!\n\nDraw a diagram and follow the steps below.\n\nThe trick is to show that the triangles DBE and \ntriangle DAF are similar.\n\nProof That triangles DBE and triangle DAF are similar.\n\nLet <ABC> donote angle ABC.\nWe need to show any two corresponding angles of these traingles are equal.\n\nFirst angle,\n<FDA> = <BDE>  because D,B,F collinear and D,E,A collinear.\n\nSecond angle,\n<DAF> = <DAC> because A,F,C collinear\n      = (pi) - <ACD> - <ADC> angles in triangle ACD\n      = (pi) - <ACD> - <ADC>  ...(1)\n\nNow, separately  manipulate,<ACD> and <ADC>.\n\n<ACD> = <ACB> + <BCD> splitting <ACD> with line CB\n      = <ACB> + <CAB>, tangent property <BCD> = <CAB> \n      = (pi) - <CBA>  angles in triangle ABC\n      = <ABE>  because C,B,E are all colinear\n\nNow, similarly\n\n<ADC> = <ADB> + <BDC> splitting <ADC> with line DB\n      = <ADB> + <BAD>, tangent property <BDC> = <BAD> \n      = (pi) - <ABD>  angles in triangle ABD\n         \nSubstituiting these two info into (1), we get,  \n\n<DAF> = (pi) - <ACD> - <ADC>\n      = (pi) - <ABE> - [ (pi) - <ABD> ]\n      = <ABD> - <ABE>\n      = <BDE> because BE splits <ABD> \n\nDone. In showing the triangle DAF and triangle DBE are similar.\n\nHence, by ratio for similar triangles,\n\n        DA/DF = DB/DE\n\nOr Rewriting we get\n \n      DA * DE = DB * DF\n\nProved.\n\nNot so difficult, was it. hehehe.\n\nRegards
467967
this is probably not a perfect answer but this site may give you some clues.\n\nhttp://www.gomath.com/Questions/question.php?question=4751\n\naltrnatively type 'prove v*v = 2gr*r/s+v0*v0-2gr' into a search engine and see what comes up.
767763
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.