The First Six, and the Eleventh and Twelfth Books of Euclid's Elements: With the Elements of Plane Trigonometry, and an Appendix in Four Books : With Notes and IllustrationsLongmans, Green, 1845 - 352 σελίδες |
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Αποτελέσματα 1 - 5 από τα 100.
Σελίδα 12
... join DC . Then , because in the triangles DBC , ACB , DB is equal to AC , and BC com- mon to both , the two sides DB ... joining CD , we prove , by means of the fourth proposition , that if the supposition were true , the triangle DBC ...
... join DC . Then , because in the triangles DBC , ACB , DB is equal to AC , and BC com- mon to both , the two sides DB ... joining CD , we prove , by means of the fourth proposition , that if the supposition were true , the triangle DBC ...
Σελίδα 15
... join BG and CF , and the straight line joining their point of intersection with A bisects the angle . The proof , which is easy , is left to exercise the learner . This problem is a particular case of the ninth proposition of the sixth ...
... join BG and CF , and the straight line joining their point of intersection with A bisects the angle . The proof , which is easy , is left to exercise the learner . This problem is a particular case of the ninth proposition of the sixth ...
Σελίδα 16
... join CG ; the straight line CG is the perpendicular required . ‡ Join CE , CF. Then , because EG is equal to GF , and CG com- * This proposition and the following contain the only two distinct cases of draw- ing a perpendicular to a ...
... join CG ; the straight line CG is the perpendicular required . ‡ Join CE , CF. Then , because EG is equal to GF , and CG com- * This proposition and the following contain the only two distinct cases of draw- ing a perpendicular to a ...
Σελίδα 19
... join ( I. post . 1 ) BE , and produce it ( I , post . 2 ) to F ; make ( I. 3 ) EF equal to BE ; and join FC . B A E HC Because ( const . ) AE is equal to EC , and BE to EF ; the two sides , AE , EB are equal to the two , CE , EF , each ...
... join ( I. post . 1 ) BE , and produce it ( I , post . 2 ) to F ; make ( I. 3 ) EF equal to BE ; and join FC . B A E HC Because ( const . ) AE is equal to EC , and BE to EF ; the two sides , AE , EB are equal to the two , CE , EF , each ...
Σελίδα 20
... joining C with the extre- mity of the produced part . The following proof is very neat and easy . From AC cut off AD equal to AB ; bisect ( I. 9 ) the angle BAC by the straight line AE , and join ED . Then , in the triangles BAE , DAE ...
... joining C with the extre- mity of the produced part . The following proof is very neat and easy . From AC cut off AD equal to AB ; bisect ( I. 9 ) the angle BAC by the straight line AE , and join ED . Then , in the triangles BAE , DAE ...
Άλλες εκδόσεις - Προβολή όλων
The First Six and the Eleventh and Twelfth Books of Euclid's Elements: With ... James Thomson, gen,James Euclides Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2016 |
Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD altitude angle ABC angle BAC angle equal BC is equal bisected centre chord circle ABC circumference cone const contained corollary 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.-To produced PROP proportional proposition pyramid radius rectangle rectilineal figure right angles Schol scholium segments semicircle sides similar solid angles square of AC straight line drawn tangent THEOR.-If third triangle ABC trigonometry triplicate ratio vertex vertical angle wherefore
Δημοφιλή αποσπάσματα
Σελίδα 130 - 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.
Σελίδα 45 - If a straight line be bisected, and produced to any point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced.
Σελίδα 2 - A rhombus is that which has all its sides equal, but its angles are not right angles.
Σελίδα 50 - 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.
Σελίδα 38 - 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.
Σελίδα 7 - If two triangles have two sides of the one equal to two sides of the...
Σελίδα 27 - 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.
Σελίδα 27 - If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles ; and the three interior angles of every triangle are equal to two right angles.
Σελίδα 15 - If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.
Σελίδα 36 - To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E.