Elementary Geometry: Plane |
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Page 29
... solution consists : ( 1 ) in showing how to use the ruler and compass so as to make the required figure ; ( 2 ) in demonstrating that the figure so constructed satisfies the prescribed conditions ; and ( 3 ) in discussing what TRIANGLES ...
... solution consists : ( 1 ) in showing how to use the ruler and compass so as to make the required figure ; ( 2 ) in demonstrating that the figure so constructed satisfies the prescribed conditions ; and ( 3 ) in discussing what TRIANGLES ...
Page 30
... solution may be possible , and under what circumstances there may be more than one solution . 68. PROBLEM 1. On a given finite line to construct an equilateral triangle . Let AB be the given line on which it is required to con- struct ...
... solution may be possible , and under what circumstances there may be more than one solution . 68. PROBLEM 1. On a given finite line to construct an equilateral triangle . Let AB be the given line on which it is required to con- struct ...
Page 33
... solution to this problem . In other words : There is only one perpendicular from a given point to a given line . Outline proof . Suppose , if possible , that OQ is a second perpendicular . Prolong OP , making PN equal to OP . Join NQ ...
... solution to this problem . In other words : There is only one perpendicular from a given point to a given line . Outline proof . Suppose , if possible , that OQ is a second perpendicular . Prolong OP , making PN equal to OP . Join NQ ...
Page 55
... solution establishes the actual existence of parallel straight lines . It shows that there is at least one line pass- ing through a given point parallel to a given line . The possibility of there being more than one is considered in the ...
... solution establishes the actual existence of parallel straight lines . It shows that there is at least one line pass- ing through a given point parallel to a given line . The possibility of there being more than one is considered in the ...
Page 62
... solution of such a problem has three divisions : ( 1 ) To make the actual construction by means of processes that ultimately involve only the drawing of straight lines and of circular arcs . ( 2 ) To prove by the use of previous ...
... solution of such a problem has three divisions : ( 1 ) To make the actual construction by means of processes that ultimately involve only the drawing of straight lines and of circular arcs . ( 2 ) To prove by the use of previous ...
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Common terms and phrases
ABCD adjacent sides altitude angle AOB angle equal angles are equal antecedents apothem base bisector bisects called central angle central line chord circumscribed circles coincide contraposite corresponding sides Definition diagonal difference divided Draw drawn equal angles equal circles equal ratios equiangular equilateral polygon equivalent figure geometry given angle given circle given line given point given polygon given ratio greater half Hence hypotenuse hypothesis inscribed interior angles intersect isosceles triangle less Let ABC line joining line-segments magnitudes measure-number mid-point mth multiple n-gon number of sides number-correspondent opposite sides parallel parallelogram perigon perimeter perpendicular PROBLEM prolonged prove quadrangle radii radius ratio compounded ratio of similitude regular polygons respectively equal rhombus right angle right triangle segments Show similar polygons similar triangles similarly solution statement straight angle straight line subtended superposable surface symmetric tangent THEOREM triangle ABC unequal vertex vertical angle
Popular passages
Page 148 - In every triangle, the square on the side subtending an acute angle is less than the sum of the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle.
Page 147 - In an obtuse-angled triangle the square on the side opposite the obtuse angle is greater than the sum of the squares on the other two sides by twice the rectangle contained by either side and the projection on it of the other side.
Page 7 - LET it be granted that a straight line may be drawn from any one point to any other point.
Page 136 - If a straight line be divided into any two parts, the square on the whole line is...
Page 195 - UPON a given straight line to describe a segment of a circle containing an angle equal to a given rectilineal angle.
Page 80 - The straight line joining the middle points of two sides of a triangle is parallel to the third side and equal to half of it 46 INTERCEPTS BY PARALLEL LINES.
Page 305 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
Page 287 - ... they have an angle of one equal to an angle of the other and the including sides are proportional; (c) their sides are respectively proportional.
Page 263 - If four magnitudes of the same kind be proportionals, they shall also be proportionals when taken alternately. Let A, B, C, D be four magnitudes of the same kind, which are proportionals, viz.
Page 32 - At a given point in a given straight line, to construct an angle equal to a given angle. Let A be the given point in the straight line AB, and O the given angle.