What is NOT a way of showing two triangles to be congruent?
363968
D, also A S S is not a way to prove uncongruence.
570600
Ok, first just so you get the figure right, draw dots n name them like i have n at the places like m showing them below\n      A                B\n      \n    E                   F     \n\n   D                      C             G\n\nNow, join AB, D-C-G, EF , A-E-D and B-F-C  also.. join\n A-F-G .. GET IT?? and you'll have AE equal to ED as well as BF equal to CF as the median of the trapezium joins the mid=points of the non-parallel sides...\n\nNOW IN triangle BFA and triangle CFG ,you have\nBF congruent to CF ----   Proved tht F midpoint of BC\nangle BFA congruent to angle CFG--vertically opposite angles\n and as AB is parallel to CD (opposite sides of a trapezoid) it is also parallel to CG ( construction was to join D,C and G , remember??\n\ntherefore, abgle ABF congruent to angle FCG --- alternate angles for parallel lines AB and CG ..\n\nSO YOU NOW HAVE triangle ABF congruent to triangle CFG --- Angle-Side-Angle Test of congruency...\n\ntherefore, AF congruent to FG --- corresponding sides for congruent triangles\n\nNOW, IN THE BIG TRIANGLE ADG,\n  E is the midpoint of side AD - by construction\n  F is the midpoint of AG  --- Proved AF congruent to FG\n \ntherefore... EF parallel to DG --- By the Midpoint Theorem which states that-- A line passing through the midpoints of any two sides of the triangle is parallel to the third side (and has half the length of the third side )\n\nSO WE'VE NOW PROVED EF // DG\n .\n. . , EF IS ALSO // TO DC --- D,C AND F LIE ON THE SAME LINE.. I.E. THEY ARE COLLINEAR\n\n SO, EF//DC\nBUT DC IS PARALLEL TO AB ---- OPPOSITE SIDES OF THE TRAPEZIUM ABCD\n \n .\n. . , EF IS ALSO PARALLEL TO AB\n\n .\n. ., WE CAN PROVE THAT ----> AB // EF // DC  (REASON- both AB and DC are parallel to the same linE EF)  \n                     \n\n .\n. . WE HAVE PROVED THAT THE MEDIAN OF THE TRAPEZIUM (in this case GE)  IS PARALLEL TO BOTH THE BASES (i.e AB and CD )
651802
The two "triangles" are not exact triangles, but rather are quadrilaterals.  The error in representation actually adds up to be one unit.  I have done the caculations in AutoCAD, and it turns out to be one square unit missing causing the extra space.
243764
im a geo student, but i havent done proofs for a long time sorry. ok\n2. false, because squares are parallelograms with 4 congruent sides and 4 right angles, while kites have 2 pairs of congruent sides and only 1 pair of congruent angles that are NOT right angles\n3. false, because a rhombus is a parallelogram with 4 congruent sides, while a kite is any thing BUT a parallelogram. enuff said\n4. false, because kites are quadrilaterals with only two pairs of congruent sides, while the quadrilateral mentioned has 3 congruent sides\n5. false, for example- do trapezoids have 4 congruent sides, and 4 right angles, and is a parallelogram (characteristic of squares)? i think NOT\n6. read above- trapezoids are not parallelograms.
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.
785502
This is a right isosceles triangle.  It's right because the Pythagorean Theorem works and it's isosceles because 2 sides are equal.
954413
Whenever I come across questions like this one, I realize two things:  (1) geometry students often feel burdened by all the strangely-named 'rules' (axioms, postulates, etc.); and (2) proofs can be intimidating!\n\nI'd like to try making this one simple, citing as few theorems as possible.  First, I will just say it "in English" for you.\n\nWhen two lines cross to form an X, you easily see at least 4 angles.  The up-down pair of angles (I'll call them v and ^) are 'vertical angles' and so are the right-left pair (I'll call them < and >).  Someone could have come up with a better name than 'vertical' ... especially since 'vertical' makes you think of "standing up straight."  Here, 'vertical' means "of the vertex."  Since the lines cross and form a vertex, the angles get called vertical.\n\nNow, why would the angles v and ^ be the same measure (congruent)?  The easiest answer is "Because they both have the same supplement."\n\nWhich supplement do they have in common?  You could say either < or >, it doesn't matter.\n\nOf course, I'm being a bit lazy by not using geometry-standard terms, so now we must label everything correctly and formally:\n\nGiven two intersecting lines, AB and CD, whose intersection is at the point X.  Prove that angle AXC is congruent to DXB.  (Hopefully your drawing will match my setup!)\n\nPROOF:\n\n(1) Angle AXC is supplementary to angle CXB.\n Reason: Angles AXC and CXB form a straight angle.\n\n(2) Angle DXB is also supplementary to angle CXB.\n Reason: (Same as Step 1)\n\n(3) Angle AXC is congruent to angle DXB.\n Reason: Two angles that are supplementary to the same third angle, are congruent.\n\nHope this sets you on your way!
986577
a) "Suppose Triangle ABC and Triangle DEF are congruent."\nb) "Suppose there exists a line segment with at least two right bisectors."\nc) "Suppose there is an isosceles triangle for which the angles opposite two of its congruent sides are not congruent."\nd) "Suppose there is a line segment with at least two midpoints."\ne) "Suppose there are two intersecting lines such that opposite angles are not congruent."\nf) "Suppose AB is not equal to CD."
151270
Then none of the angles are congruent either. :) It's just a right triangle.....
60357
It must be an isosceles trapezium: in fact the only way the two non-parallel sides can be congruent, as a consequence of the Talete's theorem, is that they form the same angle with the parallel sides.\nSo the internal angles of the quadrilateral must be congruent at couples.
139658
Diameters of the same or congruent circles are congruent. In the same or congruent\ncircles, congruent central angels have congruent arcs.\n\nIn Mathematics, congruent means:\n\nCoinciding exactly when superimposed: congruent triangles.\nOf or relating to two numbers that have the same remainder when divided by a third number. For example, 11 and 26 are congruent when the modulus is 5.
296134
There is no possible way that a triangle has a pair of parallel sides. In a pyramid, if the base is rectangular then the opposing sides of the base will be parallel but I assume that by triangular pyramid you mean the base is a triangle and each wall is a triangle...then nothing is parallel to anything.
1074757
The problem is that 30, 31, and 63 are not the sides of a valid triangle.  For any triangle with sides a,b,c, you must have\na + b > c\nb + c > a\nc + a > b\nBut you have a + b < c, so the triangle isn't valid.
987754
Actually the only way it would be possible is if you use Elliptic or Riemannian geometry, and would require a manifold.  A representation of such a construct would be similar to drawing a 4 dimensional shape in 3 or even 2 dimensions.\n\nEuclidian geometry doesn't allow triangles with a sum of all angles to be greater than 180 degrees.
164065
The trick with "proofs" like these are generally an obscured division by zero. One of the lines seems to work, though when you divide to cancel out some factors, the quantity you're dividing by could/does equal zero. It's a trick and doesn't say anything substantial about mathematics.
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.
443769
All you have to use is the fact that for any triangle, the measures of the three interior angles add to 180 degrees.  So if you have Triangles ABC and DEF, then you know:\nangleABC + angleBAC + angleBCA = 180 degrees, and\nangleDEF + angleEDF + angleEFD = 180 degrees.\n\nIf two angles are congruent, then they have the same measure in degrees.  So, if angleABC is congruent to angleDEF, and angleBAC is congruent to angleEDF, then:\nangleABC = angleDEF, and\nangleBAC = angleEDF.\n\nUsing substitution:\nangleABC + angleBAC + angleBCA = angleDEF + angleEDF + angleEFD\nangleDEF + angleBAC + angleBCA = angleDEF + angleEDF + angleEFD\nangleDEF + angleEDF + angleBCA = angleDEF + angleEDF + angleEFD.\n\nthen simplify by subtracting (angleDEF + angleEDF) from both sides:\n\nangleBCA = angleEFD.\n\ntherefore, the third angles are congruent.