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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Resultados 1-5 de 49
Página 28
... half of the parallelogram EBCA , because the diameter AB bisects ( 28. 1. ) it ; and the triangle DBC is the half of the parallelogram DBCF , because the diameter DC bisects it ; and the halves of equal things are equal ( 7. Ax ...
... half of the parallelogram EBCA , because the diameter AB bisects ( 28. 1. ) it ; and the triangle DBC is the half of the parallelogram DBCF , because the diameter DC bisects it ; and the halves of equal things are equal ( 7. Ax ...
Página 47
... half the line . Let the straight line AB be divided into two equal parts in the point C , and into two unequal parts in the point D ; the rectangle AD.DB , together with the square of CD , is equal to the square of CB , or AD.DB + ČD2 ...
... half the line . Let the straight line AB be divided into two equal parts in the point C , and into two unequal parts in the point D ; the rectangle AD.DB , together with the square of CD , is equal to the square of CB , or AD.DB + ČD2 ...
Página 48
... half the line bisected , is equal to the square of the straight line which is made up of the half and the part produced . Let the straight line AB be bisected in C , and produced to the point D ; the rectangle AD.DB together with the ...
... half the line bisected , is equal to the square of the straight line which is made up of the half and the part produced . Let the straight line AB be bisected in C , and produced to the point D ; the rectangle AD.DB together with the ...
Página 50
... half that line . " 66 SCHOLIUM . In this proposition , let the line AB be denoted by a , and the parts AC and CB by c and b ; then AD = c + 26 . Now , since a = b + c , multiplying both members by 4b , we shall have 4ab = 4b2 + 4bc ...
... half that line . " 66 SCHOLIUM . In this proposition , let the line AB be denoted by a , and the parts AC and CB by c and b ; then AD = c + 26 . Now , since a = b + c , multiplying both members by 4b , we shall have 4ab = 4b2 + 4bc ...
Página 51
... half a right angle ; and therefore the whole AEB is a right angle ; And because the angle GEF is half a right angle , and EGF a right angle , for it is equal ( 21. 1. ) to the interior and opposite angle ECB , the remaining angle EFG is ...
... half a right angle ; and therefore the whole AEB is a right angle ; And because the angle GEF is half a right angle , and EGF a right angle , for it is equal ( 21. 1. ) to the interior and opposite angle ECB , the remaining angle EFG is ...
Otras ediciones - Ver todas
Elements of Geometry: Containing the First Six Books of Euclid, with a ... John Playfair Vista de fragmentos - 1836 |
Términos y frases comunes
ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC angles equal 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 triangle DEF wherefore