In quadrilateral CDEF CD and EF are parallel, and CF and DE are parallel. Angle C is not congruent to angle D.
489938
since both sets of sides are parallel, it has to be at least a parallelogram (rules out trapezoid). And since two adjacent angles aren't the same, that rules out rectangle and square (both of which have all right angles). It can be either a rhombus or a more general parallelogram.
179545
Wow, that last person used way too much stuff.\n\nDraw in one diagonal of parallelogram ABCD, say BD.  You have congruent angles ABD and BDC, ADB and DBC because they are alternate interior angles and BD is congruent to itself.  The two triangles are congruent (ASA) which makes the parts you want congruent by CPCTC.
906923
Suppose the component vectors are a and b with magnitudes |a| and |b| and the angle between them θ\n\nThen, on completing the parallelogram of forces, the angle between the end of a and the beginning of b is (180° - θ)\n\nIf the resultant vector is R (which is the required vector), makes an angle of α between a and R, then by the sine rule\n\nsinα/|b| = sin(180° - θ)/|R|\nSo sinα = |b|sinθ/|R| (because sin(180° - θ) = sinθ\nHence α = arcsin{|b|sinθ/|R|}\n\nNote If you need |R| then by the cosine rule:\n\n|R|² = |a|² + |b|² - 2*|a|*|b|*cos(180° - θ)\n= |a|² + |b|² + 2*|a|*|b|*cosθ (since cos(180° - θ) = - cosθ)\n\nThis is for a and b in any orientation on a plane (not necessarily at right angles)\n\nIF θ = 90° then cosθ = 0 and sinθ = 1\nand these equations reduce to\n\n|R|² = |a|² + |b|²\nand \nα = arcsin{|b|/|R|} (which incidentally also equals arctan {|b|/|a|}
354533
1. True. \n2. True.\n3. True.\n4. False. A parallelogram has no symmetry at all.\n5. True.
833343
Let the triangle be ABC\n\nAB = BC\n\nso angle C = angle A (opposite angle of equal sides)\n\nAB = CA\nso angle C = angle B\n\nso all angles are same\nso equiangular
813384
I drew the picture as you described it and it looks like a parallelogram with its diagonals drawn in.  MX is half of one diagonal and MR is one side of the parallelogram.  If this is indeed the case, there is no reason why MX is congruent to MR.  So, either you need a better description of what the figure looks like, or you left out a given, such as DN is congruent to XN or something like that.
139968
false. a quadrilateral is a four sided polygon. so a square and rectangle are also quadrilaterals. a trapezoid is a four sided polygon with 2 opposite sides that are parallel to each other.
842298
If the Isosceles Triangle Theorem says, "If it's an isosceles triangle, then base angles are congruent" then the converse is "If the base angles of triangle are congruent, then the triangle is isosceles."\n\nLet's suppose we have triangle ABC, with angle B congruent to angle C.\n\nLet's draw a line from angle A to the segment BC, perpendicular to BC.  (Use the postulate: given a point and a line, there is exactly one line through the point perpendicular to the line).  The intersection of this line and segment BC is point D.\n\nNow we have two right triangles, triangle ADB and triangle ADC.  We are given angle B and angle C are congruent.  We also know that AD is congruent to AD by the reflexive property.  Therefore, ADB and ADC are congruent right triangles (perhaps AAS or LA theorems).\n\nBecause ADB and ADC are congruent right triangles, then corresponding line segments AB and AC are congruent (CPCTC).  Triangle ABC have two sides that are congruent and is therefore isosceles (definition of an isosceles triangle).\n\nIf you want the two-column proof,\nGiven: triangle ABC, angle B congruent to angle C.\nProve: triangle ABC is isosceles.\n\nStep 1: Draw line AD perpendicular to BC [Given point and line, there is only one line perpendicular to line through point.]\nStep 2: angles ADB and ADC are right angles and are congruent. [perpendicular lines intersect to form right angles; right triangles are congruent]\nStep 3: triangles ADB and ADC are right triangles. [definition of right triangles]\nStep 4: segment AD is congruent to AD [reflexive property]\nStep 5: angles B and C are congruent [Given]\nStep 6: right triangles ADB and ADC are congruent [LA or AAS]\nStep 7: line segments AB and AC are congruent [CPCTC]\nStep 8: triangle ABC is isosceles [definition of isosceles triangle]
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.
921010
I'm going to focus on the usual non-euclidean geometry due to Lobachevsky. In that geometry, there are infinitely many lines parallel to a given line and through a given point. Squares (i.e. figures with 4 equal sides and 4 equal angles) don't have right angles. There is a pentagon with all right angles. The area of a triangle can be determined by the sum of the angles in the triangle (it is proportional to the difference of 180 degrees and the sum of the angles). There is a 'natural' unit of distance obtained from angles between parallel lines and perpendicular lines.
493884
In any quadrilateral the sum of the angles is 360 degrees.  Its an axiom that has been proven.  It's similar to the axiom that all angles in a triangle add up to 180.\n\nNote: the pattern is:\n3 angles = 180\n4 angles = 360\n5 angles = 540\n6 angles = 720\n7 angles = 900\netc.\n\nAnother fact about parallelograms is that the opposite sides have the same lengths and the opposite angles have the same measure.  This also means that two consecutive angles will add up to 180 (since the other pair must be equal and the total must be 360, as noted before).
992527
E T F H L (X does not as the angles are not necessarily right angles) (I does not as it is a single line - not perpendicular to any other line)
105883
First we note that the two bases of the trapezoid are parallel and the segment joining the midpoints of the altitudes is parallel and equidistant from the bases. Therefore the vertical displacement of the midpoints of the bases from the midpoints of the altitudes is equal.\n\nIf we also contruct the segment joining the midpoints of the bases we see this intersects our other segment at its midpoint. We have now cut our quadrilateral into four triangles and it is an easy task to show that opposite triangles are congruent using SAS.
1043313
It's too cumbersome to rewrite it all here, but check out http://www.euclideanspace.com/maths/algebra/vectors/angleBetween/index.htm\n\nThat's 2D, but the formula is the same for 3D.
830003
i don;t know all of them, but here are a few:\n-rectangle\n-square\n-rhombus\n-parallelogram\n-kite\n-trapezoid\n-and closed abject with four sides!\n-cyclic quaDrilateral     \n-cyclic-inscriptable \n\ncheck out the site dr. math for more info on quadrilaterals!
317739
Let CAB = x\nCAE = x/2\nEAB = x/2\n\nADF is a right angled triangle with 90 at D.\nDFA = 90 - x/2\nCFE = 90 - x/2 (vertically opposite angles)\n\nCAE = x/2. In triangle CAE, 90 at C.\nCEF = CEA = 90 - x/2\n\nIn CEF triangle, 2 angles are equal. Sides opposite to equal angles are equal. So CE = CF.
729289
what are you supposed to prove?\n\nedit:\nyou mean why the three vertices of an equilateral triangle are concyclic? i am unable to make out theproblem