Books 3-9The University Press, 1908 |
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Αποτελέσματα 1 - 5 από τα 99.
Σελίδα 9
... less than the greater : which is impossible . Therefore the straight line joined from A to B will not fall outside the circle . Similarly we can prove that neither will it fall on the circumference itself ; therefore it will fall within ...
... less than the greater : which is impossible . Therefore the straight line joined from A to B will not fall outside the circle . Similarly we can prove that neither will it fall on the circumference itself ; therefore it will fall within ...
Σελίδα 12
... less to the greater : which is impossible . Therefore AC , BD do not bisect one another . Therefore etc. PROPOSITION 5 . Q. E. D. If two circles cut one another , they will not have the same centre . For let the circles ABC , CDG cut ...
... less to the greater : which is impossible . Therefore AC , BD do not bisect one another . Therefore etc. PROPOSITION 5 . Q. E. D. If two circles cut one another , they will not have the same centre . For let the circles ABC , CDG cut ...
Σελίδα 15
... less than the angle CFE . Therefore EM is less than EC , and therefore than EB . Hence the point B in which FB meets the circle 111. 7 ] 15 PROPOSITION 7.
... less than the angle CFE . Therefore EM is less than EC , and therefore than EB . Hence the point B in which FB meets the circle 111. 7 ] 15 PROPOSITION 7.
Σελίδα 16
... less angles at that point with the straight line containing it and the centre is the more instructive and useful of the two , since it is such lines drawn in any manner to the circle from the point which are immediately useful in the ...
... less angles at that point with the straight line containing it and the centre is the more instructive and useful of the two , since it is such lines drawn in any manner to the circle from the point which are immediately useful in the ...
Σελίδα 17
... less than DH , AE is not greater but less than AF Moreover this cannot be proved by the same method as before . For , while we can prove that GE > HF , GA > AH , H E B we cannot make any inference as to the comparative length of AE , AF ...
... less than DH , AE is not greater but less than AF Moreover this cannot be proved by the same method as before . For , while we can prove that GE > HF , GA > AH , H E B we cannot make any inference as to the comparative length of AE , AF ...
Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD angle ABC angle BAC antecedent Aristotle base bisected centre circle ABC circumference construction continued proportion corresponding sides cube number definition diameter drawn enunciation equal angles equiangular equimultiples Euclid Eutocius ex aequali four magnitudes geometrical geometrical progression given circle given straight line greater ratio greatest common measure Heiberg hypothesis Iamblichus joined less mean proportional numbers measures the number multiple multitude Nicomachus odd number parallel parallelogram pentagon polygon Porism prime number Proclus Prop proper fraction proposition PROPOSITION 13 proved rect rectangle rectangle contained rectilineal figure reductio ad absurdum remaining angle right angles segment semicircle similar and similarly similar plane numbers Simson solid numbers square number subtracted taken Theon Theon of Smyrna theorem touches the circle triangle ABC unit VIII δὲ καὶ πρὸς τὸ τοῦ
Δημοφιλή αποσπάσματα
Σελίδα 34 - EQUAL straight lines in a circle are equally distant from the centre ; and those which are equally distant from the centre, are equal to one another.
Σελίδα 65 - If a straight line touch a circle, and from the point of contact a chord be drawn, the angles which this chord makes with the tangent are equal to the angles in the alternate segments.
Σελίδα 37 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...
Σελίδα 234 - Prove that similar triangles are to one another in the duplicate ratio of their homologous sides.
Σελίδα 64 - From this it is manifest that if one angle of a triangle be equal to the other two it is a right angle, because the angle adjacent to it is equal to the same two ; (i.
Σελίδα 209 - IN a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another.
Σελίδα 90 - EF at right angles (9. 1.) to AB, AC ; DF, EF produced meet one another : for, if they do not meet, they are parallel, wherefore AB, AC, which are at right angles to them, are parallel ; which is absurd : let them meet in F, and join FA ; also, if the point F be not in BC, join BF, CF : then, because AD is equal to DB, and DF common, and at right angles to AB, the base AF is equal (4.
Σελίδα 80 - In a given circle to place a straight line, equal to a given straight line which is not greater than the diameter of the circle. Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle.
Σελίδα 212 - ... be parallel to the remaining side of the triangle. Let DE be drawn parallel to BC, one of the sides of the triangle ABC : BD is to DA, as CE to EA. Join BE, CD ; Then the triangle BDE is equal to the triangle CDE*, * «.i.
Σελίδα 95 - In the same manner, it may be demonstrated that the straight lines EC, ED, are each of them equal to EA or EB ; therefore the four straight lines EA, EB, EC, ED, are equal to one another ; and the circle described from the centre E, at the distance of one of them, shall pass through the extremities of the other three, and be described about the square ABCD.