given c is the perpendicular bisector de AB prove  ANGLE ABC IS ISOCELES?
828988
angles are never isoceles... may be you mean the triangle? \nand also, what is C, where is it located? '
798688
in triangles ABE and BEC\n<AEB=<BEC=90*\nAB=BC (given)\n<ABD=<BCE(given\ntherefore the triangles are congruent\ntherefore BD=EC\nbut EC=AE as in an isosceles triangle the altitude bisects the base\nso BD=AE (cpct)\nhence proved
140559
Using the abc method… \n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nFe-->ax=2c------eq.1\nO-->ya+4b=12c+2d+e----eq.2\nH-->2b=2e----------eq.3\nS-->b=3c+d----------eq.4\n\nLet a=1\n\nFrom eq.1\nax=2c\nx/2=2/2c\nx/2=c-----eq.5\n\nForm eq.3\n2b=2e\n2/2b=2/2e\ne=b-----eq.6 \n\nFrom eq.2\nYa+4b=12c+2d+e\n1y+4b=12(x/2) +2d+b (Form eq.5&6)\nY+3b=6x+2d\n2d=y-6x+3b\nd=(y+3b-6x)/2------------eq.7\n\nFrom eq.4\nb=3c+d (but c=x/2)\nb=3(x/2) +d\nReplacing eq.4 by eq.7 we will get\nb=3x/2+(y+3b-6x) ½\nb= (3x+y+3b-6x) ½\n2[b=(y+3b-3x) ½]\n2b=y+3b-3x\n3x-y=b-------eq.8\n\nReplacing eq.6 by eq.8\ne=b\ne=3x-y\n\nReplacing eq.7 by eq.8\nd=(y+3b-6x)/2\nd=[y+3(3x-y)-6x]/2\nd=[y+9x-3y-6x]/2\nd= (3x-2y)/2\n\nFrom the above calculation we know that:\na=1, b=3x-y, c=x/2, d=(3x-2y)/2 & e=3x-y. \n\n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nWhen replacing the values we’ll get:\n[1]FexOy+[3x-y]H2SO4 -->[x/2]Fe2(SO4)3+[(3x-2y)/2]SO2 +[3x-y]H2O
928672
let the base of the isosceles triangle be x\nheight of the triangle=h\ngiven the radius of the circle is 4\n(4-h)^2+(x/2)^2=4^2\n16-8h+h^2+x^2/4=16\n4h^2-32h+x^2=0\nx=rt(32-4h^2)\narea=1/2*x*h\n=1/2*h[2rt(8-h^2)]\ndA/dh=0 for max\nafter finding h sub to get x
310544
Distance formula=sq.root of{(x2-x1)sq.+(y2-y1)sq.}\n\nDistance AB=(from formula) sq.rt. of 89\nDistance BC=(from formula) 10\nDistance AC=(from formula) sq.rt. of 89\n\nFrom above-AB=AC.\nTwo sides are equal hence ABC is an isoceles triangle.
938489
I'd cheat and take the cross product to get a vector normal to the plane of those two vectors.  Then to prove that it's a basis, you must prove that they are linearly independent.\n\nOr\n\nnotice the zero component in v1?  It lies in the x-z plane.  So, pick a vector in the x-y or y-z plane.  Since v2 is not in any of those planes you should be OK.  Also, vectors with zero components are easy to visualize and use.
776798
Every side of this triangle has a pair of ends and all the three pairs are cyclic. ie, if AB,BC & CA are the three sides of the given triangle , then the side AB has the ends (4,5),(1,1), BC has the ends  (1,1)and (-2,5) & CA has the pair of ends (4,5),(-2,5).\nBy simple inspection of these pairs we can find that the points C and A are at the same height from the abcissa hence parallel to it. By this we can conclude easily that the length of CA=6 units.\nFurther, to be an isocelous triangle in this case the point B should be on the mid line of CA, and it is normal to CA. The mid point of CA is three units away from its ends C&A. As CA is parallel to absissa we can clearly conclude that any point between (1,0)&(1,5) will be the vertex of an isosceles triangle. Here the point B is (1,1) on the said line. Thus proved that the given points form an isosceles triangle.\nWe alredy know that CA = 6\nAnd the other sides are equal to sqrt(4-1)^2 +(5-1)^2 = 5 units\nNow, we can find the area of this triangle by Heron's Formula \nie sqrt[8(8-6)(8-5)^2] =12 sq units
728779
Sounds like youve got a nasty integral. Your going to have to integrate a polar vector in cartesian coordinates. Write the B field in cartesian and then integrate one side of the square.\n\nB.ds= miu (0) * I is true. I forget which theorem but I think its stokes theorem from vector calculus. It doesnt matter which shape you choose. The symmetry of the circle allows a simple integral because there is no dot product for the B.s since they are always parrallel. Choosing a square just makes your life horrible. If you correctly write the B field and then use the vector identity:\n\ncos theta = a.b/(|a||b|)\n\nto find the angle for the dot product you will get the same answer. It's just hard to set up the problem.\n\nIm sorry dude, good luck.
389433
If you're asking how to do it: Just take the total number of degrees of the angle, divide by two, and there you go, you know the measurements of the 2 equal angles.\nanswered by Fading Away...\n\nAngle bisector Theorem.\n\nIf you are a Geometry student who knows constructions, locate a radi from your vertex and draw an arc. Choose another radii on that new arc that you just drew(That is more than half the arc) from the point where your newly drawn arc intersects the segment. Draw another arc. Repeat process on opposite side. Draw bisectore from vertex through intersection of two arcs.\n\nHope that helps!\nAnswered by Emily\n\n\n\nThis was asked and answered 13 hours ago, click this link.\nhttp://answers.yahoo.com/question/;_ylt=Al6hLACuaPLouhbP7BmwxcIjzKIX?qid=1006041024466
109605
Just use Cosinus Theorem: c^2=a^2+b^2-2abCos(gamma)\n1.\nc=sqrt(4+64-2*2*8*sqrt(3)/2))\nc=6.34\n\n2. Is not a right Triangle.. but anyway lets find the angle A:\n22^2=15^2+33^2-2*15*33*Cos(A)\nCos(A)=830/990\nA=ArcCos(83/99)\nA=33°1' 48.54"
729594
Rhombus and square (c)\n\n\nThis can be proved using geometry, analytic geometry, etc.  But it is the case.  Surprisingly, a general reactangle and parallelograms don't have perpendicular diagonals.\n\nGeneral kites do, and a rhombus is a kite with all equal sides.  A square can be thought of as a type and rhombus and rectangle, or just a rhombus when each angle is 90 degrees.  Then it makes sense that it would have perpendicular diagonals.
67852
this is difficult to do without drawing.  but here's the gist of it.\n\n1. <a + <b + <c + <d = 360  (definition of quadrilateral)\n2. <a + <b + <a + <b = 360  (since a = c and b = d)\n3. 2(a + b) = 360\n4. a + b = 180\n5. <a and <b are supplementary angles\n\nnow extend the line AB and include a new point X, so you have a picture that looks like:\n\nX\n\nA   D\n\n\nB   C\n\n<XAD and <BAD are supplementary since XAB is a line\n\nabove we proved that <ABC and <BAD are also supplementary\n\nso <XAD + <BAD = 180\nand <ABC + <BAD = 180\nwhich means <XAD = <ABC\n\nso now we can conclude that AD || BC\nbecause the corresponding angles <XAD and <ABC are congruent.\n\nyou can use this same logic to show that AB || DC\n\nhope this makes sense..
496924
Well, you know that each corner of the octagon lies on the circle (by definition). So, if A and B are the names of two adjacent corners, and C the center of the circle, you also know that AC=14.15, and BC=14.15. \n\nAnd you also know that a regular octagon will have its corners equally spaced around the circle. Since a circle is 360 degrees, the AC/BC angle is 360/8 = 45 degrees.\n\nFrom there you can easily calculate the length of AB using simple trigonometry, which is probably the goal of the exercise :-)
93089
This seems like a homework problem, so I'll give a few hints.\n\nThe arclength is radius times the angle subtended.\n\nTo find the angle subtended, consider the isosceles triangle created inside the circle where the chord is the hypotenuse and the radius of the circle is the two other legs. \n\nSince the isosceles triangle created is not a right triangle you need to use a specific formula to find the angle "C" subtended by the chord.
856774
Angle BDE = 30 degrees\n\nFirst, since you have an isoceles triangle, with A being 20 degrees, that means that angles B and C will be 80 degrees (20 + 80 + 80 = 180).\n\nCBD is 50 degrees, so CBD is 30.\nSimilarly:\nBCE is 40 degrees, so DCE is also 40 degrees.\n\nYour two lines (BD and CE) cross at a point, let's call that F.\nIf you check the measurements of the triangles around it, you'll see that the four angles around F are all 90 degrees.\n\nTriangle CBF = 40-50-90\nTriangle CDF = 40-50-90\nTriangle BEF = 30-60-90\n\nSince CBF and CDF are congruent triangles, it follows that BEF and DEF must be congruent triangles.\n\nTherefore BDE must be 30 degrees.
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
552905
The AREA of an isosceles triangle is found by taking 1/2 the sum of the long and short parallel sides, times the height. One way to compare this with the parallelogram is to take two copies, turn one 180 degrees and glue them along the equal (nonparallel) edges. \nYou now have a larger parallelogram - with one side composed of the sum of the original short and long parallel edges. Use the usual formula for the area of this parallelogram, then divide by 2 (since it is composed of two equal copies).