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 páginas |
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Página 17
... cut , are together equal to four right angles . COR . 2. And consequently that all the angles made by any number of lines meeting in one point , are together equal to four right angles . " C Book I. a 10. 1 . PROP . XVI . OF EUCLID . 17.
... cut , are together equal to four right angles . COR . 2. And consequently that all the angles made by any number of lines meeting in one point , are together equal to four right angles . " C Book I. a 10. 1 . PROP . XVI . OF EUCLID . 17.
Página 86
... consequently the other 32. 1. two angles BAD , ABD , are equald to a right angle : But ABF is like- wise a right angle : therefore the E angle ABF is equal to the angles BAD , ABD : Take from these equals the common angle ABD ...
... consequently the other 32. 1. two angles BAD , ABD , are equald to a right angle : But ABF is like- wise a right angle : therefore the E angle ABF is equal to the angles BAD , ABD : Take from these equals the common angle ABD ...
Página 103
... consequently the three angles BDA , DBA , BCD , are equal to one another ; and because the angle DBC is equal to the angle BCD , the side BD is equal to the side DC ; " 6. 1 . but BD - was made equal to CA , therefore also CA is equal ...
... consequently the three angles BDA , DBA , BCD , are equal to one another ; and because the angle DBC is equal to the angle BCD , the side BD is equal to the side DC ; " 6. 1 . but BD - was made equal to CA , therefore also CA is equal ...
Página 135
... consequently adding NM to both , LM is greater than NP : K Therefore , if GH be greater than KO , LM is greater than NP . In like manner it may be shown , that if GH be equal to KO , LM is equal to NP ; and if less , less . And in the ...
... consequently adding NM to both , LM is greater than NP : K Therefore , if GH be greater than KO , LM is greater than NP . In like manner it may be shown , that if GH be equal to KO , LM is equal to NP ; and if less , less . And in the ...
Página 142
... consequently F the least . AB , together with F , are greater than CD , together with E. Take AG equal to E , and CH equal to F : Then because as AB is to CD , so is E to F , and that AG is equal to E , and CH equal to F , AB is to CD ...
... consequently F the least . AB , together with F , are greater than CD , together with E. Take AG equal to E , and CH equal to F : Then because as AB is to CD , so is E to F , and that AG is equal to E , and CH equal to F , AB is to CD ...
Términos y frases comunes
ABC is given ABCD AC is equal altitude angle ABC angle BAC base BC bisected Book XI centre circle ABC circumference common logarithm cone cylinder demonstrated described diameter draw 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 sides BA similar sine solid angle solid parallelopipeds square of BC straight line AB straight line BC tangent THEOR third triangle ABC vertex wherefore
Pasajes populares
Página 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.
Página 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.
Página 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.
Página 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.
Página 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...
Página 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.
Página 1 - A straight line is that which lies evenly between its extreme points.
Página 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.
Página 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.
Página 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...