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 TrigonometryCollins & Hannay, 1860 - 333 páginas |
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Página 74
... joins the points F , G shall pass through the point of contact . For , if not , let it pass otherwise , if possible , FCDG , and join FA , AG and because F is the centre of the circle ABC , AF is equal to FC : Also because G is the ...
... joins the points F , G shall pass through the point of contact . For , if not , let it pass otherwise , if possible , FCDG , and join FA , AG and because F is the centre of the circle ABC , AF is equal to FC : Also because G is the ...
Página 85
... join BC , EF : the straight line BC is equal to the straight line EF . Take ( 1. 3. ) K , L , the centres of the circles , and join BK , KC , EL , LF : and because the arch BGC is equal to the arch EHF , the angle BKC is equal ( 27. 3 ...
... join BC , EF : the straight line BC is equal to the straight line EF . Take ( 1. 3. ) K , L , the centres of the circles , and join BK , KC , EL , LF : and because the arch BGC is equal to the arch EHF , the angle BKC is equal ( 27. 3 ...
Página 86
... Join AB , and bisect ( 19. 1. ) it in C ; from the point C draw CD at right angles to AB , and join AD , DB ; the arch ADB is bisected in the point D. D Because AC is equal to CB , and CD common to the triangle ACD , BCD , the two sides ...
... Join AB , and bisect ( 19. 1. ) it in C ; from the point C draw CD at right angles to AB , and join AD , DB ; the arch ADB is bisected in the point D. D Because AC is equal to CB , and CD common to the triangle ACD , BCD , the two sides ...
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Términos y frases comunes
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 and similar 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 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 side BC sine solid angle solid parallelopipeds spherical angle spherical triangle straight line AC THEOR third touches the circle triangle ABC triangle DEF wherefore