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 TrigonometryE. Duyckinck, and George Long, 1824 - 333 σελίδες |
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Σελίδα 19
... semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter . XV . Rectilineal figures are those which are contained by straight lines . XVI . Trilateral figures , or triangles , by three ...
... semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter . XV . Rectilineal figures are those which are contained by straight lines . XVI . Trilateral figures , or triangles , by three ...
Σελίδα 63
... semicircle BHF and produce DE to H , and join GH . Therefore , because the straight line BF is divided into two equal parts in the point G , and into two unequal in the point E , the rect- angle BE.EF , together with the square of EG ...
... semicircle BHF and produce DE to H , and join GH . Therefore , because the straight line BF is divided into two equal parts in the point G , and into two unequal in the point E , the rect- angle BE.EF , together with the square of EG ...
Σελίδα 80
... semicircle , and join BF , FD : And because the angle BFD is at the cen- tre , and the angle BAD at the circumfer- ence , both having the same part of the circumference , viz . BCD , for their base ; therefore the angle BFD is double ...
... semicircle , and join BF , FD : And because the angle BFD is at the cen- tre , and the angle BAD at the circumfer- ence , both having the same part of the circumference , viz . BCD , for their base ; therefore the angle BFD is double ...
Σελίδα 81
... semicircle ; and the angles in it BAC , BEC are equal , by the first case : For the same reason , because CBED is greater than a semicir- cle , the angles CAD , CED are equal ; Therefore the whole angle BAD is equal F E to the whole ...
... semicircle ; and the angles in it BAC , BEC are equal , by the first case : For the same reason , because CBED is greater than a semicir- cle , the angles CAD , CED are equal ; Therefore the whole angle BAD is equal F E to the whole ...
Σελίδα 82
... the segment ABC is a semicircle . Next let the angles ABD , BAD be unequal ; at the point A , in the straight line AB make ( 23 . 1. ) the angle BAE equal to the angle ABD , and produce BD if neces A B D B D A C E A B 82 ELEMENTS.
... the segment ABC is a semicircle . Next let the angles ABD , BAD be unequal ; at the point A , in the straight line AB make ( 23 . 1. ) the angle BAE equal to the angle ABD , and produce BD if neces A B D B D A C E A B 82 ELEMENTS.
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ABC is equal ABCD altitude angle ABC angle ACB angle BAC angle EDF arch AC base BC bisected centre circle ABC circumference cosine cylinder demonstrated diameter draw equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given straight line greater hypotenuse inscribed join less Let ABC Let the straight line BC magnitudes meet multiple opposite angle parallel parallelepipeds parallelogram perpendicular polygon prism PROB produced proportionals proposition Q. E. D. COR Q. E. D. PROP radius ratio rectangle contained rectilineal figure remaining angle segment semicircle shewn side BC sine solid angle solid parallelepipeds spherical angle spherical triangle straight line AC THEOR third touches the circle triangle ABC triangle DEF wherefore