Write out the proof proving that vertical angles are congruent.?
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!
628857
Not a, since the congruent sides could be adjacent sides.\n\nNot b, because that would be a trapezoid.\n\nDefinitely c - there's a way to prove this using properties of parallel lines, interior angles, and triangles.\n\nMaybe d, since congruent diagonals means that your angles could be right angles, but it also could mean that you have an isosceles trapezoid.\n\nBisecting diagonals always means that your quadrilateral is a parallelogram.  Congruent diagonals can also yield a parallelogram, albeit a specialized one - a rectangle.  Now if you have diagonals that both bisect each other and are congruent, then you have a square.
350854
64 proofs\nexplanation\nProof #1\nThis is probably the most famous of all proofs of the Pythagorean proposition. It's the first of Euclid's two proofs (I.47). The underlying configuration became known under a variety of names, the Bride's Chair likely being the most popular.\n\n    \n\nThe proof has been illustrated by an award winning Java applet written by Jim Morey. I include it on a separate page with Jim's kind permission.\n\nFirst of all, ABF = AEC by SAS. This is because, AE = AB, AF = AC, and\n\n  BAF = BAC + CAF = CAB + BAE = CAE.\n \n\nABF has base AF and the altitude from B equal to AC. Its area therefore equals half that of square on the side AC. On the other hand, AEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Thus the area of AEC equals half that of the rectangle AELM. Which says that the area AC2 of the square on side AC equals the area of the rectangle AELM.\n\nSimilarly, the are BC2 of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB.\n\nThe configuration at hand admits numerous variations. B. F. Yanney and J. A. Calderhead (Am Math Monthly, v.4, n 6/7, (1987), 168-170 published several proofs based on the following diagrams\n\n   \n\n\n\nProof #2\n    \n\nWe start with two squares with sides a and b, respectively, placed side by side. The total area of the two squares is a2+b2.\n\n    \n\nThe construction did not start with a triangle but now we draw two of them, both with sides a and b and hypotenuse c. Note that the segment common to the two squares has been removed. At this point we therefore have two triangles and a strange looking shape.\n\n    \n\nAs a last step, we rotate the triangles 90o, each around its top vertex. The right one is rotated clockwise whereas the left triangle is rotated counterclockwise. Obviously the resulting shape is a square with the side c and area c2.\n\n(A variant of this proof is found in an extant manuscript by Thâbit ibn Qurra located in the library of Aya Sofya Musium in Turkey, registered under the number 4832. [R. Shloming, Thâbit ibn Qurra and the Pythagorean Theorem, Mathematics Teacher 63 (Oct., 1970), 519-528]. ibn Qurra's diagram is similar to that in proof #27. The proof itself starts with noting the presence of four equal right triangles surrounding a strangenly looking shape as in the current proof #2. These four triangles correspond in pairs to the starting and ending positions of the rotated triangles in the current proof. This same configuration could be observed in a proof by tesselation.)\n\n\n\nProof #3\n    \n\nNow we start with four copies of the same triangle. Three of these have been rotated 90o, 180o, and 270o, respectively. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.\n\n    \n\nThe square has a square hole with the side (a-b). Summing up its area (a-b)2 and 2ab, the area of the four triangles (4·ab/2), we get\n\n  c2 = (a-b)2+2ab \n  = a2-2ab+b2+2ab \n  = a2+b2 \n \n\n\n\nProof #4\n    \n\nThe fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side (a+b) and a hole with the side c. We can compute the area of the big square in two ways. Thus\n\n  (a + b)2 = 4·ab/2 + c2  \n\nsimplifying which we get the needed identity.\n\n\n\nProof #5\n\nThis proof, discovered by President J.A. Garfield in 1876 [Pappas], is a variation on the previous one. But this time we draw no squares at all. The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - (a+b)/2·(a+b). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab/2 + ab/2 + c·c/2. As before, simplifications yield a2+b2=c2.\n\nTwo copies of the same trapezoid can be combined in two ways by attac
646629
If the angle from the base is the same in the reflection as the angle from the base of your fan isosceles triangle then the angles are congruent.
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You need to start from the definintions of x < y and x > y.\n\nThey probably go like this\nif x < y, then there exists 0 < z such that x+z = y.\nif x > y, then there exists 0 < w such that x = y+w.\n\nPartial proof:\nif x < y, and x = y, then for some z, 0 < z, x+z = y.\nsince you have x = y, too, y + z = y, z = 0. That contradicts 0 < z. Proving the other two parts is similar.\nx+z = y,
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a^2 + b^2 = c^2\n\nPythagorean theorem
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The angles all add up to 180 degrees.\n\nIf you know the lengths of all the sides you can use the "Law of Tangents" to find all the angles.  This is advanced geometry and Trigonometry, so rather than type, I am including a link to a pretty good explanation.\n\nhttp://answers.google.com/answers/threadview?id=38131
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Then none of the angles are congruent either. :) It's just a right triangle.....
8996
Well, I agree with Rosie. You can do proofs really fast if you have a very clear understanding of all the required theorems & formulae. But for this you need to really really practice - and the time won't be far away when the only time that will be required will be the time you take to write the proof! I was very poor in this subject - but I was so determined that I would sit long hours and try to prove it on my own. Believe it or not I would spent 2 or 3 hours for trying to prove certain problems  i.e without taking anybody's help. It was very tough initially - sometimes i was not able to do it - i would go to sleep at night with the problem in mind - honestly i would always get it right the next morning. With gradual practice i could do it in a jiffy! But my dear don't forget H.W.Longfellow - "The heights of great men reached and kept, Were not obtained by sudden flight, But they, while their companions slept Were toiling upward in the night."
170974
Prove that for  u,v, and w, p1=s(u)+t(v)+r(w), i.e that you can create a linear combination of vectors such that v=a1v1+a2v2+...+anvn.
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Can't u prove it ?\nuse simple geometrical angle laws.\n\nif we have a triangle : \nupside-- the inclined plane\ndownside--- the horizontal\nright side--- the gravitational force (vertical)\n\nDRAW IT........ as u know \n\nand then draw the perpendicual component on the plane     \n\nuse this: sum of the two angles inside the triangle    =180-90=90\n\nsum of angles at the gravitional point= 180-90=90 also\n\nu will prove the equality
454426
i have a great logical reasoning for that.... here goes:\n1. there are two 2s, so you overlap them, making just one two. \n2. take the plus sign out for a second.\n2 1/2. the plus sign consists of 2 sticks, one horizantal and one vertical. \n3. the vertical came become a one and place it on the left of the 2. ( it makes 12)\n4. the other horizantal line can be used as a minus sign. \n5. Now, in the equal sign, it consists of two lines, both horizantal. \n5 1/2. if you leave one, and make the other diagonal, and attach in together, it makes a seven. \n6. so.. so far, it's is like this: 12-7=?\n7. ultimately, the answers five\nso there you have it, it sounds a little weird and strange, but it's really impossible to have a logical reasoning for 2+2=5... hehe
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It is a tough Q - coz it is NOT right angle...\nThe best solution would be to name the angles, LHS Triangle as angle x, y z,,,,,     RHS triangle as  x', y', z'\n\nThe actual proof will be long, but the basic idea is\nSum of the Angles of a Tri = 180\nand the angles in straight line = 180\n\nx + y + z =180\n\nx' + y' + z' = 180\n\nz + y' = 180\n\ny' = 180-z                                                 \n\nx = x'\n\nx' + 180-z + z' = 180\nx + z' = z\n\nTherefore, z is larger Angle than x  (or x')\n\nSide opposite to larger angle is larger
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Your statement is incorrect.  It reads:\n\n"In the image below, there are right angles. How would I prove that they're right angles without a protractor in any triangle?"\n\nActually, it should read:\n\n"In the image below, there are some angles."\n\nAnything else is pure conjecture.  (That means a guess.)\n\nIf you would like us to say something about those angles, you need to tell us how the figure was constructed.  After you tell us that, and ask your question more clearly, your answer will be obtained in short order.\n\nedit...\n\nYou have added some text in your extra text file that doesn't help at all:\n\n"I know that these green points are right angles..."\n\nSo you think a point can also be an angle.  It will be a long time before you get someone (who knows and understands geometry) to agree with that.\n\nIf you start over and explain what you want to know, in correct terms, you may have a chance at getting an answer.
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1.  Opposite angles in a rhombus are congruent.\n2.  Adjacent angles in a rhombus are supplmentary.\n\n<LMK is half of one of the smaller angles, and <KJM is one of the larger angles.  So <LMJ will be 72, and since <LMJ and <KJM are supplementary, <KJM is 108.
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Congruent.  \n\nComplementary angles are angles whose sum is 90˚.  So for an angle to be complementary to another anlge, it must equal 90˚ minus the size of the other angle.  If to two angles are complimentary to the same angle, they both equal 90˚ minus the size of that other angle, and since that other angle doesn't change, the sizes of the two angles must be equal, which means they're congruent.\n\nA few people have said the angles are "equal."  While they have the right idea, angles actually have no numerical value, and therefore can't be equal to each other.  The correct term for geometic objects (like angles) that have the same size and shape (i.e. are "equal") is congruent.
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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.