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, Edlemnts of Plane and Spherical TrigonometryW.E. Dean, 1836 - 311 páginas |
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Página 28
... triangle ABC is equal to the triangle DBC . PROP . XXXII . B THEOR . C Triangles upon equal bases , and between the same parallels , are equivalent to one another . Let the triangles ABC , DEF be upon equal bases BC , EF , and be- tween ...
... triangle ABC is equal to the triangle DBC . PROP . XXXII . B THEOR . C Triangles upon equal bases , and between the same parallels , are equivalent to one another . Let the triangles ABC , DEF be upon equal bases BC , EF , and be- tween ...
Página 116
... triangles AHG , AGB , ABC are all equal ( 32. 1. ) ; Therefore , whatever multiple the base HC is of the base BC , the same multiple is the triangle AHC of the triangle ABC . For the same reason , whatever the base LC is of the base CD ...
... triangles AHG , AGB , ABC are all equal ( 32. 1. ) ; Therefore , whatever multiple the base HC is of the base BC , the same multiple is the triangle AHC of the triangle ABC . For the same reason , whatever the base LC is of the base CD ...
Página 123
... triangle ABC may be proved to be equiangular to the triangle DEF , as in the first case . PROP . VIII . THEOR . In a right angled triangle if a perpendicular be drawn from the right angle to the base ; the triangles on each side of it ...
... triangle ABC may be proved to be equiangular to the triangle DEF , as in the first case . PROP . VIII . THEOR . In a right angled triangle if a perpendicular be drawn from the right angle to the base ; the triangles on each side of it ...
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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 demonstrated diameter divided draw Prob 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 solid parallelopiped spherical angle spherical triangle square straight line BC THEOR third touches the circle triangle ABC wherefore