The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh and Twelfth ... Also the Book of Euclid's Data, in Like Manner CorrectedWingrave and Collingwood, 1816 - 528 sider |
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Side
... demonstrated , and not assumed ; and therefore , though this were a true proposition , it ought to have been demonstrated . But indeed , this Proposition , which makes the 10th definition of the 11th Book , is not true universally ...
... demonstrated , and not assumed ; and therefore , though this were a true proposition , it ought to have been demonstrated . But indeed , this Proposition , which makes the 10th definition of the 11th Book , is not true universally ...
Side 9
... demonstrated . PROP . V. THEOR . THE angles at the base of an isosceles triangle are equal to one another ; and if the equal sides be produced , the angles upon the other side of the base / shall be equal . Le ABC be an isosceles ...
... demonstrated . PROP . V. THEOR . THE angles at the base of an isosceles triangle are equal to one another ; and if the equal sides be produced , the angles upon the other side of the base / shall be equal . Le ABC be an isosceles ...
Side 10
... demonstrated , that the whole angle ABG is equal to the whole ACF , the parts of which , the angles CBG , BCF are also equal ; the remaining angle ABC is therefore equal to the remaining angle ACB , which are the angles at the base of ...
... demonstrated , that the whole angle ABG is equal to the whole ACF , the parts of which , the angles CBG , BCF are also equal ; the remaining angle ABC is therefore equal to the remaining angle ACB , which are the angles at the base of ...
Side 14
... demonstrated , that two straight lines cannot have a common segment . If it be possible , let the two straight lines ABC , ABD have the segment AB common to both of them . From the point B draw BE at rightangles to AB ; and because ABC ...
... demonstrated , that two straight lines cannot have a common segment . If it be possible , let the two straight lines ABC , ABD have the segment AB common to both of them . From the point B draw BE at rightangles to AB ; and because ABC ...
Side 17
... demonstrated , that no other can be in the same straight line with it but BD , which therefore is in the same straight line with CB . Wherefore , if at a point , & c . Q. E. D. PROP . XV . THEOR . IF two straight lines cut one another ...
... demonstrated , that no other can be in the same straight line with it but BD , which therefore is in the same straight line with CB . Wherefore , if at a point , & c . Q. E. D. PROP . XV . THEOR . IF two straight lines cut one another ...
Almindelige termer og sætninger
ABC is given AC is equal altitude angle ABC angle BAC base BC bisected centre circle ABCD circumference common logarithm cone cylinder demonstrated described diameter drawa drawn equal angles equiangular equimultiples Euclid excess fore given angle given in magnitude given in position given in species given magnitude given point given ratio given straight line gnomon greater join less Let ABC logarithm multiple opposite parallel parallelogram AC perpendicular point F polygon prism proportionals proposition Q. E. D. PROP radius ratio of AE rectangle CB rectangle contained rectilineal figure remaining angle right angles segment side BC similar sine solid angle solid parallelopipeds square of BC straight line AB straight line BC tangent THEOR third triangle ABC vertex wherefore
Populære passager
Side 41 - If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.
Side 180 - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.
Side 166 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides. Let AC, CF be equiangular parallelograms having the angle BCD equal to the angle ECG ; the ratio of the parallelogram AC to the parallelogram CF is the same with the ratio which is compounded •f the ratios of their sides. DH Let BC, CG be placed in a straight line ; therefore DC and CE are also in a straight line (14.
Side 2 - A rhomboid, is that which has its opposite sides equal to one another, but all its sides are not equal, nor its angles right angles.
Side 105 - The first of four magnitudes is said to have the same ratio to the second, which the third has to the fourth, when any equimultiples whatsoever of the first and third being taken, and any equimultiples whatsoever of the second and fourth ; if the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth...
Side 79 - The angle in a semicircle is a right angle; the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.
Side 1 - A straight line is that which lies evenly between its extreme points.
Side 149 - 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 23 - That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
Side 83 - Wherefore from the given circle ABC has been cut off the segment BAC, containing an angle equal to the given angle DQEP PROP. XXXV. THEOR. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other. Let the...