Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle and the Geometry of Solids ; to which are Added Elements of Plane and Spherical Trigonometry |
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Σελίδα 45
Let the paralellogram ABCD and the triangle EBC be upon the same base BC and between the same parallels BC , AE ; the parallelogram ABCD A DE is double of the triangle EBC . Join AC ; then the triangle ABC is equal ( 37. 1. ) ...
Let the paralellogram ABCD and the triangle EBC be upon the same base BC and between the same parallels BC , AE ; the parallelogram ABCD A DE is double of the triangle EBC . Join AC ; then the triangle ABC is equal ( 37. 1. ) ...
Σελίδα 47
Let the paralellogram ABCD and the triangle EBC be upon the same base BC and between the same parallels BC , AE ; the parallelogram ABCD A D E is double of the triangle EBC . Join AC ; then the triangle ABC is equal ( 37. 1. ) ...
Let the paralellogram ABCD and the triangle EBC be upon the same base BC and between the same parallels BC , AE ; the parallelogram ABCD A D E is double of the triangle EBC . Join AC ; then the triangle ABC is equal ( 37. 1. ) ...
Σελίδα 48
Let ABCD be a parallelogram of which the diameter is AC ; let EH , FG be the parallelograms about AC , that is , through which AC passes , and let BK , KD be the other parallelograms , which make up the A H D whole figure ABCD and are ...
Let ABCD be a parallelogram of which the diameter is AC ; let EH , FG be the parallelograms about AC , that is , through which AC passes , and let BK , KD be the other parallelograms , which make up the A H D whole figure ABCD and are ...
Σελίδα 49
It is required to describe a parallelogram equal to ABCD , and having an angle equal to E. G Join DB , and describe ( 42. 1. ) the OF GEOMETRY . BOOK . I. 49 V ...
It is required to describe a parallelogram equal to ABCD , and having an angle equal to E. G Join DB , and describe ( 42. 1. ) the OF GEOMETRY . BOOK . I. 49 V ...
Σελίδα 50
And because the triangle ABD is equal to the parallelogram HF and the triangle DBC to the parallelogram GM , the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM ; therefore the parallelogram KFLM has been ...
And because the triangle ABD is equal to the parallelogram HF and the triangle DBC to the parallelogram GM , the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM ; therefore the parallelogram KFLM has been ...
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ABC is equal ABCD altitude angle ABC angle BAC arch base bisected Book called centre circle circle ABC circumference coincide common cylinder definition demonstrated described diameter difference divided double draw drawn equal equal angles equiangular Euclid exterior angle extremity fall fore four fourth given given straight line greater half inscribed interior join less Let ABC magnitudes manner meet multiple opposite parallel parallelogram pass perpendicular plane polygon prism produced PROP proportionals proposition proved radius ratio reason rectangle contained rectilineal figure right angles segment shewn sides similar sine solid square straight line taken tangent THEOR thing third touches triangle ABC wherefore whole