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 16
... exterior angle of a triangle ( 9. 1. ) is greater than the interior and opposite angle , the exterior angle BDC of the triangle CDE is greater than CED ; for the same reason ... angle EDG equal to the angle BAC : and make DG 16 ELEMENTS.
... exterior angle of a triangle ( 9. 1. ) is greater than the interior and opposite angle , the exterior angle BDC of the triangle CDE is greater than CED ; for the same reason ... angle EDG equal to the angle BAC : and make DG 16 ELEMENTS.
Página 119
... exterior angle of the triangle . Let the exterior angle CAE , of any triangle ABC , be bisected by the straight line AD which meets the base produced in D ; BD is to DC , as BA to AC . E Through C draw CF parallel to AD ( Prob . 13. 1 ...
... exterior angle of the triangle . Let the exterior angle CAE , of any triangle ABC , be bisected by the straight line AD which meets the base produced in D ; BD is to DC , as BA to AC . E Through C draw CF parallel to AD ( Prob . 13. 1 ...
Página 292
... angle AGH is equal to EGB , because these are vertical , and it has also been shewn to be equal to GHD , therefore ... exterior angle of a triangle be bisected , and also one of the interior and opposite , the angle contained by the ...
... angle AGH is equal to EGB , because these are vertical , and it has also been shewn to be equal to GHD , therefore ... exterior angle of a triangle be bisected , and also one of the interior and opposite , the angle contained by the ...
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Términos y frases comunes
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