A Treatise on Elementary Geometry: With Appendices Containing a Collection of Exercises for Students and an Introduction to Modern Geometry |
From inside the book
Page 86
... given circles intersect each other , only the exterior tangents are possible ... circle is wholly within the other , there is no solution . PROPOSITION XLI . - PROBLEM . 94. To inscribe a circle in a given triangle . A Let ABC be the given ...
... given circles intersect each other , only the exterior tangents are possible ... circle is wholly within the other , there is no solution . PROPOSITION XLI . - PROBLEM . 94. To inscribe a circle in a given triangle . A Let ABC be the given ...
Page 117
... triangle ABC , AB2 + BC2 = 2AE2 + 2BE3 , and in the triangle ADC , CD2 + DA2 ... bisect each other , and the distance EF is zero . PROPOSITION XX . - THEOREM . 65. In any triangle , the product of two sides is equal to the product of the ...
... triangle ABC , AB2 + BC2 = 2AE2 + 2BE3 , and in the triangle ADC , CD2 + DA2 ... bisect each other , and the distance EF is zero . PROPOSITION XX . - THEOREM . 65. In any triangle , the product of two sides is equal to the product of the ...
Page 118
... triangle ABC , AD the perpendicular upon BC , AE the di- ameter of the circumscribed circle ... triangles AEC , ABD , are similar , and give AB : AE AD : AC ... Let AD bisect the angle A of the triangle ABC ; then , ABX AC = DB × DC + DA ...
... triangle ABC , AD the perpendicular upon BC , AE the di- ameter of the circumscribed circle ... triangles AEC , ABD , are similar , and give AB : AE AD : AC ... Let AD bisect the angle A of the triangle ABC ; then , ABX AC = DB × DC + DA ...
Page 146
... triangle ABO , we have AB1 = OA2 + OB1 20A2 , whence AB = OA . V / 2 , by which the side of the inscribed square can be computed , the radius being given . = PROPOSITION V. - PROBLEM . 14. To inscribe a regular hexagon in a given circle ...
... triangle ABO , we have AB1 = OA2 + OB1 20A2 , whence AB = OA . V / 2 , by which the side of the inscribed square can be computed , the radius being given . = PROPOSITION V. - PROBLEM . 14. To inscribe a regular hexagon in a given circle ...
Page 147
... inscribed equilateral triangle is of one right angle = 120 ° . PROPOSITION VI . - PROBLEM . 17. To inscribe a regular decagon in a given circle . Suppose the problem solved , and let ABC .... L , be a regular inscribed decagon . Join AF ...
... inscribed equilateral triangle is of one right angle = 120 ° . PROPOSITION VI . - PROBLEM . 17. To inscribe a regular decagon in a given circle . Suppose the problem solved , and let ABC .... L , be a regular inscribed decagon . Join AF ...
Other editions - View all
A Treatise on Elementary Geometry: With Appendices Containing a Collection ... William Chauvenet No preview available - 2016 |
A Treatise on Elementary Geometry: With Appendices Containing a Collection ... William Chauvenet No preview available - 2015 |
Common terms and phrases
ABCD AC² adjacent angles altitude apothem base bisects centre of similitude chord circumference circumscribed coincide common cone Corollary cylinder Definition denote diagonals diameter dicular diedral angle distance divided draw edges equally distant equilateral equivalent exterior angle faces figure find the locus frustum given circles given plane given point given straight line hence homologous hypotenuse indefinitely inscribed angle isosceles Let ABC measure medial lines middle point number of sides one-half opposite sides parallelogram parallelopiped pendicular perimeter perpen perpendicular plane MN plane passed polar pole polyedral angle polyedron prism PROPOSITION pyramid quadrilateral radical axis radii radius rectangle regular polygon respectively right angles right triangle Scholium secant segment similar sphere spherical polygon square straight line drawn straight line joining surface symmetrical tangent tetraedron theorem three given triangle ABC triedral vertex vertices volume
Popular passages
Page 128 - The area of a rectangle is equal to the product of its base and altitude.
Page 348 - Three lines are in harmonical proportion, when the first is to the third, as the difference between the first and second, is to the difference between the second and third ; and the second is called a harmonic mean between the first and third. The expression 'harmonical proportion...
Page 19 - The perpendicular is the shortest line that can be drawn from a point to a straight line.
Page 79 - To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line ; it is required to draw a straight line through the point A, parallel to the straight hue BC.
Page 29 - The sum of the three angles of any triangle is equal to two right angles.
Page 175 - If a straight line is perpendicular to each of two straight lines at' their point of intersection, it is perpendicular to the plane of those lines.
Page 263 - The sum of the angles of a spherical triangle is greater than two and less than six right angles ; that is, greater than 180° and less than 540°. (gr). If A'B'C' is the polar triangle of ABC...
Page 219 - A truncated triangular prism is equivalent to the sum of three pyramids, whose common base is the base of the prism and whose vertices are the three vertices of the inclined section.
Page 197 - A right prism is a prism •whose lateral edges are perpendicular to the planes of the bases.
Page 127 - Any two rectangles are to each other as the products of their bases by their altitudes. Let R and R...