The Elements of Plane and Solid Geometry: With Chapters on Mensuration and Modern Geometry |
From inside the book
Results 1-5 of 84
Page 16
... drawing perpendiculars , equal angles , squares , etc. accurately with ruler and compass , and by the methods ... draw , on paper , triangles of various shapes , and cut them out ; then cut off two of the angles , and place them ...
... drawing perpendiculars , equal angles , squares , etc. accurately with ruler and compass , and by the methods ... draw , on paper , triangles of various shapes , and cut them out ; then cut off two of the angles , and place them ...
Page 18
... draw ( Post . G H D E L F K 1 ) the lines GD , GF . The triangle GDF is the required triangle . Proof . Because DG is a radius of the circle GHK , it is equal ( Def . 30 ) to B ; and because FG is a radius of GKL , it is equal to C ...
... draw ( Post . G H D E L F K 1 ) the lines GD , GF . The triangle GDF is the required triangle . Proof . Because DG is a radius of the circle GHK , it is equal ( Def . 30 ) to B ; and because FG is a radius of GKL , it is equal to C ...
Page 25
... draw a straight line at right angles to a given straight line from a given point in it . Let AB be a straight line , and C a given point in it . It is required to draw from Ca straight line at right angles to AB . Take any point D , in ...
... draw a straight line at right angles to a given straight line from a given point in it . Let AB be a straight line , and C a given point in it . It is required to draw from Ca straight line at right angles to AB . Take any point D , in ...
Page 26
... draw a straight line perpendicular to a given straight line of unlimited length , from a given point without it . Let AB be a straight line , and C a point without it . It is required to draw from C a perpendicular to AB . Take any ...
... draw a straight line perpendicular to a given straight line of unlimited length , from a given point without it . Let AB be a straight line , and C a point without it . It is required to draw from C a perpendicular to AB . Take any ...
Page 28
... Draw ( I. 11 ) BE at right angles to CD . Then CBE = R and EBD = R. Also ABC = CBE + ABE = R + ABE , and ABD = EBD – ABE = R – AВЕ . ... ( Ax . 2 ) ABC + ABD = 2R . Proposition 16 . E A C B D Theorem . If at a point in a straight line ...
... Draw ( I. 11 ) BE at right angles to CD . Then CBE = R and EBD = R. Also ABC = CBE + ABE = R + ABE , and ABD = EBD – ABE = R – AВЕ . ... ( Ax . 2 ) ABC + ABD = 2R . Proposition 16 . E A C B D Theorem . If at a point in a straight line ...
Other editions - View all
The Elements of Plane and Solid Geometry: With Chapters on Mensuration and ... Isaac Sharpless No preview available - 2016 |
The Elements of Plane and Solid Geometry: With Chapters on Mensuration and ... Isaac Sharpless No preview available - 2017 |
Common terms and phrases
A-BCD ABē ABCD ACē altitude angle ABC angle ACB angle BAC apothem bisect centre of similitude chord circle ABC circumference cone Corollary cylinder decagon describe diagonals diameter divided draw equal angles equiangular feet figure four right angles frustum given circle given straight line greater Hence inscribed interior angles intersect isosceles Let ABC line joining meet middle point multiplied number of sides opposite angles parallelogram parallelopiped pass perimeter perpendicular plane pole polyedron prism produced Prop proportional Proposition 12 Proposition 13 pyramid quadrilateral radical axis radii radius rectangle contained regular polygon right angles Scholium segment semicircle similar slant height solid solid angle sphere spherical angle spherical triangle square surface symmetrical tangent Theorem three angles three sides triangle ABC vertex
Popular passages
Page 53 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.
Page 81 - On a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle. Let AB be the given straight line, and...
Page 31 - Any two angles of a triangle are together less than two right angles.
Page 128 - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.
Page 15 - AXIOMS. 1. Things which are equal to the same thing are equal to one another. 2. If equals be added to equals the wholes are equal. 3. If equals be taken from equals the remainders are equal.
Page 82 - To cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle. Let ABC be the given circle, and D the given rectilineal angle ; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the given angle D.
Page 62 - If a straight line be divided into two equal, and also into two unequal parts ; the squares on the two unequal parts are together double of the square on half the line, and of the square on the line between the points of section.
Page 166 - A be a solid angle contained by any number of plane angles BAC, CAD, DAE, EAF, FAB: these together are less than four right angles. Let the planes...
Page 15 - LET it be granted that a straight line may be drawn from any one point to any other point. 2. That a terminated straight line may be produced to any length in a straight line.
Page 120 - Similar polygons may be divided into the same number of similar triangles, having the same ratio to one another that the polygons have ; and the polygons have to one another the duplicate ratio of that which their homologous sides have.