Books 3-9The University Press, 1908 |
Inni boken
Resultat 1-5 av 78
Side 35
... square on AF is equal to the square on CG , for AF is equal to CG ; therefore the square on FE which remains is equal to the square on EG , therefore EF is equal to EG . But in a circle straight lines are said to be equally distant from ...
... square on AF is equal to the square on CG , for AF is equal to CG ; therefore the square on FE which remains is equal to the square on EG , therefore EF is equal to EG . But in a circle straight lines are said to be equally distant from ...
Side 36
... square on EF is equal to the square on EG , for EF is equal to EG ; therefore the square on AF which remains is equal to the square on CG ; therefore AF is equal to CG . And AB is double of AF , and CD double of CG ; therefore AB is ...
... square on EF is equal to the square on EG , for EF is equal to EG ; therefore the square on AF which remains is equal to the square on CG ; therefore AF is equal to CG . And AB is double of AF , and CD double of CG ; therefore AB is ...
Side 37
... square on EK , the square on BH is greater than the square on FK . It may be that Euclid would have regarded this as too complicated an inference to make without explanation or without an increase in the number of his axioms . But , on ...
... square on EK , the square on BH is greater than the square on FK . It may be that Euclid would have regarded this as too complicated an inference to make without explanation or without an increase in the number of his axioms . But , on ...
Side 65
... square on AD [ 1. 47 ] , that is , to the square on BE . And the squares on CO , BO are equal to the square on BC . Therefore , by addition , the squares on AO , BO , CO , DO are equal to the squares on EB , BC , i.e. to the square on ...
... square on AD [ 1. 47 ] , that is , to the square on BE . And the squares on CO , BO are equal to the square on BC . Therefore , by addition , the squares on AO , BO , CO , DO are equal to the squares on EB , BC , i.e. to the square on ...
Side 72
... square on EG is equal to the square on GC ; [ II . 5 ] Let the square on GF be added ; therefore the rectangle AE , EC together with the squares on GE , GF is equal to the squares on CG , GF . But the square on FE is equal to the ...
... square on EG is equal to the square on GC ; [ II . 5 ] Let the square on GF be added ; therefore the rectangle AE , EC together with the squares on GE , GF is equal to the squares on CG , GF . But the square on FE is equal to the ...
Vanlige uttrykk og setninger
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
Populære avsnitt
Side 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.
Side 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.
Side 37 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...
Side 234 - Prove that similar triangles are to one another in the duplicate ratio of their homologous sides.
Side 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.
Side 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.
Side 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.
Side 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.
Side 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.
Side 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.