Euclid's Elements of Geometry, Bøker 1-6;Bok 11Henry Martyn Taylor The University Press, 1895 - 657 sider |
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Side xvi
... similarly situate figures . The chief difficulty with respect to the additions which have been made to Book VI was the immense number of known theorems from which a selection had to be made . I have attempted by means of two or three ...
... similarly situate figures . The chief difficulty with respect to the additions which have been made to Book VI was the immense number of known theorems from which a selection had to be made . I have attempted by means of two or three ...
Side 30
... Similarly it can be proved that AC coincides with DF . Therefore the triangle ABC coincides with the triangle DEF , and is equal to it in all respects . Wherefore , if two triangles & c . EXERCISES . 1. If a quadrilateral have two pairs ...
... Similarly it can be proved that AC coincides with DF . Therefore the triangle ABC coincides with the triangle DEF , and is equal to it in all respects . Wherefore , if two triangles & c . EXERCISES . 1. If a quadrilateral have two pairs ...
Side 46
... the angle CEA is equal to the angle DEB . Similarly it may be proved that the angle CEB is equal to the angle AED . Wherefore , if two straight lines & c . EXERCISES . 1. If the diagonals of a quadrilateral bisect 46 BOOK I.
... the angle CEA is equal to the angle DEB . Similarly it may be proved that the angle CEB is equal to the angle AED . Wherefore , if two straight lines & c . EXERCISES . 1. If the diagonals of a quadrilateral bisect 46 BOOK I.
Side 48
... Similarly it can be proved that the angle BCG , which is made by producing AC and is equal to the angle ACD , is greater than the angle ABC . Wherefore , an exterior angle & c . EXERCISES . 1. Only one perpendicular can be drawn to 48 ...
... Similarly it can be proved that the angle BCG , which is made by producing AC and is equal to the angle ACD , is greater than the angle ABC . Wherefore , an exterior angle & c . EXERCISES . 1. Only one perpendicular can be drawn to 48 ...
Side 53
... Similarly it can be proved that AO is equal to BO ; therefore BO is equal to Co. Next because in the triangles BOD , COD , ( Prop . 4. ) BO is equal to CO and BD to CD and OD is common , the triangles are equal in all respects ...
... Similarly it can be proved that AO is equal to BO ; therefore BO is equal to Co. Next because in the triangles BOD , COD , ( Prop . 4. ) BO is equal to CO and BD to CD and OD is common , the triangles are equal in all respects ...
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Vanlige uttrykk og setninger
ABCD ADDITIONAL PROPOSITION AE is equal angle ABC angle ACB angle BAC angular points bisected bisectors centre of similitude chord circle ABC circumscribed circle coincide CONSTRUCTION Coroll cut the circle describe a circle diagonals diameter equal angles equiangular equilateral triangle Euclid EXERCISES figure fixed point given circle given point given straight line given triangle greater harmonic range hypotenuse inscribed circle intersect isosceles triangle Let ABC locus magnitudes meet middle points opposite sides pairs parallel parallelepiped parallelogram perpendicular plane angles polygon PROOF Prop quadrilateral radical axis radius rectangle contained regular polygon required to prove respectively rhombus right angles right-angled triangle shew side BC Similarly solid angle sphere square on AC straight line drawn straight line joining tangent tetrahedron theorem triangle ABC triangles are equal trihedral angle twice the rectangle vertex vertices Wherefore
Populære avsnitt
Side 70 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Side 218 - The angle which an arc of a circle subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.
Side 279 - If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together •with the square...
Side 148 - If a straight line be divided into any two parts, the squares on the whole line, and on one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square on the other part.
Side 137 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.
Side 372 - To find a mean proportional between two given straight lines. Let AB, BC be the two given straight lines ; it is required to find a mean proportional between them. Place AB, BC in a straight line, and upon AC describe the semicircle ADC, and from the point B draw (9.
Side 78 - ... the same side together equal to two right angles ; the two straight lines shall be parallel to one another.
Side 303 - To inscribe, an equilateral and equiangular pentagon in a given circle. Let ABCDE be the given circle. It is required to inscribe an equilateral...
Side 420 - PROPOSITION 5. The locus of a point, the ratio of whose distances from two given points is constant, is a circle*.
Side 300 - To inscribe a circle in a given square. Let ABCD be the given square ; it is required to inscribe a circle in ABCD.