The Elements of Euclid, books i. to vi., with deductions, appendices and historical notes, by J.S. Mackay. [With] Key

Εξώφυλλο

Αναζήτηση στο βιβλίο

Επιλεγμένες σελίδες

Άλλες εκδόσεις - Προβολή όλων

Συχνά εμφανιζόμενοι όροι και φράσεις

Δημοφιλή αποσπάσματα

Σελίδα 142 - UPON a given straight line to describe a segment of a circle containing an angle equal to a given rectilineal angle.
Σελίδα 44 - Equal triangles on the same base, and on the same side of it, are between the same parallels.
Σελίδα 117 - THEOREM. // two circumferences intersect each other, the distance between their centres is less than the sum, and greater than the difference, of their radii.
Σελίδα 244 - ... equal to the rectangle AE, BD together with the rectangle CE, BD ; that is, equal to the rectangle contained by BD and the sum of AE and EC. But the sum of AE and EC is greater than AC (I. 20); therefore the rectangle AB, DC, together with the rectangle BC, AD is greater than the rectangle AC, BD. 33. If the rectangle contained by the diagonals of a quadrilateral be equal to the sum of the rectangles contained by the opposite sides, a circle can be described round the quadrilateral. This is the...
Σελίδα 136 - To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line.
Σελίδα 137 - With A as centre and AB as radius describe a circle. with B as centre and BA as radius describe a circle cutting the former one in two points.
Σελίδα 118 - And whenever the triangle CAD can be constructed, it is plain that the circles described from the centres C and D, will cut each other in A and B. COR. 1. Hence, if the distance between the centres of two circles be greater than the sum of their radii, the two circles will not intersect each other. COR. 2. Hence, also, if the distance between the centres be less than the difference of the radii, the two circles will not cut each other. For, AC+CD>AD; therefore, CD>AD— AC ; that is, any side of...
Σελίδα 142 - ... join DB. Then AD is equal to the given straight line. And the angle ACB is equal to the sum of the angles CDB and CBD (I. 32), that is, to twice the angle CDB (I. 5). Therefore the angle ADB is half of the angle in the given segment. Hence we have the following synthetical solution. Describe on AB a segment of a circle containing an angle equal to half the angle in the given segment. With A as centre, and a radius equal to the given straight line, describe a circle. Join A with a point of intersection...
Σελίδα 83 - No. 13.) 22. If three points be taken on the sides of a triangle such that the sums of the squares of the alternate segments taken cyclically are equal, the perpendiculars to the sides of the triangle at these points are concurrent. (TG de Oppel, "Analysis Triangulorurn,
Σελίδα 41 - ... these triangles are equal to twice as many right angles as the figure has sides. But all the angles of these triangles are equal to the interior angles of the figure, viz., ABC, BCD, &c., together with the angles at the point F, which are equal (I.

Πληροφορίες βιβλιογραφίας