complementary angles, vertical angles, equilateral triangles, isosceles triangles, obtuse triangles, right triangles, parallelogram, scalene triangles, supplementary angles, and a trapezoid.
138216
U can see the whole world as Geo-metric figures. The Design of a house by civil Engineer is full of Geometry. Followed by lands Planned to use for building purpose and for Ariel view. And with the help of ice in winters if u go to the north and south pole u can find natural and Man made designs which r Geometric figures which u have mentioned above.\n\n  Geometry is not the different types of angles and triangles it includes figures like cone, circle, cylinders , Pentagon and u can go on  Infinity
201089
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
146249
A quadrilateral is a four-sided polygon with four angles. There are many kinds of quadrilaterals. The five most common types are the parallelogram, the rectangle, the square, the trapezoid, and the rhombus. \n The sum of the angles of a quadrilateral is 360 degrees. \n Angle CB  would equal 2 because it is congruent to angle AD.
517758
The answer comes from the definitions of the terms used.  Let's go through them:\n\nA trapezoid is a quadrilateral with one pair of opposite sides parallel.  The parallel sides are called the BASES.\n\nThe other two sides are called the LEGS.\n\nThe MEDIAN of a trapezoid is a LINE SEGMENT connecting the mid points of the legs of a trapezoid.  Hence the median is // to the bases.\n\nIt is the same for a parallelogram but since both opposites sides are // it doesn't matter which ones form the base and which ones form the legs.  The median will be // to the base you chose.\n\nHope that helps.
453821
Arrange the four points into a triangular pyramid - one whose four faces are all equilateral triangles.
438771
No picture to be found.  \n\nAre M, N, O, P all on the same line?  \n\nIf so, NOQ is isoceles, which makes MNQ and QOP congruent angles.  Since two sides and the included angle uniquely define the third side, so MQ and QP are congruent. and MQP is isoceles.
540122
a triangle with angles 30, 60,90\nhas sides in ratio 1: (SR 3) : 2\nThe equilateral triangle can be formed from\ntwo 30,60,90 triangles.\neach of those can be formed from 3 smaller 30,60,90 triangles\nso the equilateral triangle can be formed from six 30,60,90 triangles.\nIf radius of circle is 1 unit,\neach of the six small 30,60,90 triangles has \nsides (1/2) ((SR 3)/2) 1\nso the equilateral triangle has side SR(3) units\nso if the equilateral triangle has side x, the circle has radius x /(SR(3)),\nand the circle circumference is 2 pi x /((SR 3))
403876
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.
139562
now, if you add that it has 4 sides and all right angles, you probably can pass that off as the definition of a rectangle.\n\notherwise you just have a parallelogram.
56501
an isoceles triangle pointing up\nand another pointing down right on top of it
539528
Regular Polygon - all angles are equal and all sides are the same length. Regular polygons are both equiangular and equilateral.
292985
could be circle because it doesn't have any sides or angles\n\ncould be rectangle because it is not proportional (side-wise.. and only if the triangle is equilateral)\n\ncould be triangle because if all shapes are placed right-side and divided into four equal parts, like a cross-hair, the triangle is the only one without equal portions.
166350
Triangles can be classified in two ways:\n\nBy the relative lengths of their sides:\n\n1.  Equilateral:  all sides are of equal length\n2.  Isosceles:  at least two sides are equal\n3.  Scalene:  all sides have different lengths \n   \nBy the size of their largest internal angle:\n\n1.  Right:  has one 90°internal angle\n2.  Obtuse:  has one internal angle larger than 90° \n3.  Acute:  all internal angles are smaller than 90°
361467
It is possible only when the given triangle is isosceles or equilateral. It's also possible for right-angled triangles but only when the given angle isn't 90 degrees.
209620
Use the double angle identities.
531363
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!
554490
Here's my answer:\n\nhttp://mudandmuck.com/str2/triangles.JPG\n\nYou didn't say write them out in English, you said describe.  This picture describes the solution.