The first book of Euclid's Elements, simplified, explained and illustrated, by W. Trollope1847 |
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Resultat 1-5 av 22
Side 4
... common boundary of the two superficies AL , KC ; and , as it partakes of the breadth of neither , it has length only without breadth . So again the point K is the common extremity of two lines AK , BK ; and , partaking of the length of ...
... common boundary of the two superficies AL , KC ; and , as it partakes of the breadth of neither , it has length only without breadth . So again the point K is the common extremity of two lines AK , BK ; and , partaking of the length of ...
Side 10
... common intersection , which has been already seen to be impossible ( Note on Definition 6 ) . 11. All right angles are equal to one another . ( See Definition 10. ) 12. If a straight line meets two straight lines , 10 EUCLID.- -BOOK I.
... common intersection , which has been already seen to be impossible ( Note on Definition 6 ) . 11. All right angles are equal to one another . ( See Definition 10. ) 12. If a straight line meets two straight lines , 10 EUCLID.- -BOOK I.
Side 23
... common to both , ... the two DB , BC = the two AC , CB , each to each ; and the DBC = ACB ( Hyp . ) ; .. base DC = base AB , and the area of DBC area of △ ACB : i.e. the < = > , which is absurd . is 2. Hence AB is not unequal to AC : i ...
... common to both , ... the two DB , BC = the two AC , CB , each to each ; and the DBC = ACB ( Hyp . ) ; .. base DC = base AB , and the area of DBC area of △ ACB : i.e. the < = > , which is absurd . is 2. Hence AB is not unequal to AC : i ...
Side 24
... common to the △ FBC , GCB , .. the two sides FB , BC = GC , CB , each to each ; and the contained / FBC = Z GCB ( Hyp . ) ; ... the base FC = base GB , and the BCF ( Prop . IV . ) 2. Prove that But BCG = GCF = LFBG . = LCBG . CBF ( Hyp ...
... common to the △ FBC , GCB , .. the two sides FB , BC = GC , CB , each to each ; and the contained / FBC = Z GCB ( Hyp . ) ; ... the base FC = base GB , and the BCF ( Prop . IV . ) 2. Prove that But BCG = GCF = LFBG . = LCBG . CBF ( Hyp ...
Side 30
... - Because AD AE , * B D F and AF is common to AS DAF , EAF , .. the two sides DA , AF = EA , AF , each to each ; and the base DF base EF ( Const . ) ; .. DAF = EAF . ( Prop . VIII . ) line , the DBE = EBA ( Def . 10 28 EUCLID . - BOOK I.
... - Because AD AE , * B D F and AF is common to AS DAF , EAF , .. the two sides DA , AF = EA , AF , each to each ; and the base DF base EF ( Const . ) ; .. DAF = EAF . ( Prop . VIII . ) line , the DBE = EBA ( Def . 10 28 EUCLID . - BOOK I.
Vanlige uttrykk og setninger
ABCD adjacent angle contained base BC bisect CD Prop coincide Const CONST.-In CONST.-Join CONST.-Let DEMONST.-Because DEMONST.-For demonstration diam diameter draw EBCF ENUN ENUN.-If ENUN.-Let ABC ENUN.-To ENUN.-To describe equal sides equilateral Euclid EUCLID'S ELEMENTS exterior four rt given point given straight line interior and opposite interior opposite isosceles join Let ABC line be drawn line drawn meet opposite angles opposite sides parallel parallelogram perpendicular Post PROB produced Proposition proved rectilineal figure rhombus right angles side BC square take any pt THEOR THEOR.-If Theorem trapezium trapezium ABCD vertical Wherefore XXIX XXXI XXXII XXXIV XXXVIII
Populære avsnitt
Side 58 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal...
Side 24 - Upon the same base, and on the same side of it, there cannot be two triangles, that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity, equal to one another.
Side 34 - If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.
Side 6 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the other is called a perpendicular to it.
Side 109 - If the square described upon one of the sides of a triangle, be equal to the squares described upon the other two sides of it; the angle contained by these two sides is a right angle.
Side 9 - Parallel straight lines are such as are in the same plane, and which being produced ever so far both ways, do not meet.
Side 99 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
Side 49 - If, from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than, the other two sides of the triangle, but shall contain a greater angle. Let...
Side 104 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.
Side 6 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.