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 TrigonometryW.E. Dean, 1836 - 311 páginas |
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Página 62
... chord AB , and at the same time in the perpendicular EF bisecting the chord BC ( Cor . 1. Th . 3. 3. ) , must be at the intersection of these perpendiculars ; so that , as there is but one centre , there can be but one circumference ...
... chord AB , and at the same time in the perpendicular EF bisecting the chord BC ( Cor . 1. Th . 3. 3. ) , must be at the intersection of these perpendiculars ; so that , as there is but one centre , there can be but one circumference ...
Página 65
... chord AD , would coincide with those of the chord BD ; these chords are therefore equal : hence , the angle ACD is equal to the angle BCD ( Th . V. B. I. ) . COR . 1. It follows , moreover , that equal angles at the centre are sub ...
... chord AD , would coincide with those of the chord BD ; these chords are therefore equal : hence , the angle ACD is equal to the angle BCD ( Th . V. B. I. ) . COR . 1. It follows , moreover , that equal angles at the centre are sub ...
Página 170
... chord of one - sixth of the circumference , if P = the perpendicular from C on the chord of one - twelfth of the circumference , P will be a mean proportional between AH ( 8. 1. Sup . ) and AC + CG , and P2 = AH ( AC + CG ) = 500 ...
... chord of one - sixth of the circumference , if P = the perpendicular from C on the chord of one - twelfth of the circumference , P will be a mean proportional between AH ( 8. 1. Sup . ) and AC + CG , and P2 = AH ( AC + CG ) = 500 ...
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ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC angles equal arc AC base BC bisected centre chord circle ABC circumference cosine cylinder definition demonstrated diameter divided equal and similar equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given straight line greater Hence hypotenuse inscribed join less Let ABC Let the straight magnitudes meet multiple opposite angle parallel parallelogram parallelopiped perpendicular polygon prism PROP proposition quadrilateral radius ratio rectangle contained rectilineal figure remaining angle right angled triangle SCHOLIUM segment semicircle shewn side BC sine solid angle spherical angle spherical triangle square straight line BC THEOR third touches the circle triangle ABC triangle DEF wherefore