The First Six and the Eleventh and Twelfth Books of Euclid's Elements: With Notes and Illustrations, and an Appendix in Five BooksA. & C. Black, 1837 - 390 sider |
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Side 11
... reason . When a proposition is demonstrated by supposing one figure to be applied to another , it is said to be proved by the method of superposition . PROP . V. THEOR . THE angles at the base Book I. ] 11 OF EUCLID .
... reason . When a proposition is demonstrated by supposing one figure to be applied to another , it is said to be proved by the method of superposition . PROP . V. THEOR . THE angles at the base Book I. ] 11 OF EUCLID .
Side 13
... reason . In other respects this proof is perfectly satisfactory ; and on account of its ease and sim- plicity , it may perhaps be preferred by those who are reading the Elements for the first time . This proposition may also be proved ...
... reason . In other respects this proof is perfectly satisfactory ; and on account of its ease and sim- plicity , it may perhaps be preferred by those who are reading the Elements for the first time . This proposition may also be proved ...
Side 24
... reason , the exterior angle CEB of the triangle ABE is greater than BAC ; and it has been de- monstrated that the angle BDC is greater than CEB ; much more then is the angle BDC greater than BAC . Therefore , if from a point , & c ...
... reason , the exterior angle CEB of the triangle ABE is greater than BAC ; and it has been de- monstrated that the angle BDC is greater than CEB ; much more then is the angle BDC greater than BAC . Therefore , if from a point , & c ...
Side 37
... reason , the parallelogram EG is equal to the same parallelogram EBCH ; because it is on the same base EH , and between the same parallels EH , BG . Therefore also the parallelogram AC is equal ( I. ax . 1. ) to EG . Wherefore paral ...
... reason , the parallelogram EG is equal to the same parallelogram EBCH ; because it is on the same base EH , and between the same parallels EH , BG . Therefore also the parallelogram AC is equal ( I. ax . 1. ) to EG . Wherefore paral ...
Side 40
... reason , the triangles KGC , KFC are equal . Then , because the triangle AEK is equal to AHK , and KGC to KFC , the triangles AEK , KGC are together equal ( I. ax . 2. ) to the triangles AHK , KFC taken together . But the whole triangle ...
... reason , the triangles KGC , KFC are equal . Then , because the triangle AEK is equal to AHK , and KGC to KFC , the triangles AEK , KGC are together equal ( I. ax . 2. ) to the triangles AHK , KFC taken together . But the whole triangle ...
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The First Six and the Eleventh and Twelfth Books of Euclid's Elements: With ... Euclid Ingen forhåndsvisning tilgjengelig - 2016 |
Vanlige uttrykk og setninger
ABCD altitude angle ABC angle BAC angle equal BC is equal bisected centre chord circle ABC circumference cone const contained cylinder describe a circle diagonal diameter divided draw equal angles equal to AC equiangular equilateral Euclid exterior angle fore fourth given circle given point given ratio given straight line greater half Hence hypotenuse inscribed join less Let ABC magnitudes manner multiple opposite parallel parallelepiped parallelogram perpendicular polygon polyhedron prism PROB produced PROP proportional proposition pyramid radius rectangle rectilineal figure right angles Schol segments semicircle sides similar similar triangles solid angles square of AC straight lines drawn tangent THEOR third triangle ABC triplicate ratio vertex vertical angle wherefore
Populære avsnitt
Side 94 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Side 53 - If a straight line be divided into any two parts, the squares of the whole line and of one of the parts are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts at the point C : the squares of AB, BC shall be equal to twice the rectangle AB, BC, together with the square of AC.
Side 143 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Side 4 - A rhombus is that which has all its sides equal, but its angles are not right angles.
Side 57 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part.
Side 138 - IF a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those produced, proportionally; and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle...
Side 43 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.
Side 32 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 40 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
Side 36 - PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another...