The Elements of Euclid for the Use of Schools and CollegesMacmillan, 1872 - 400 Seiten |
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Seite viii
... less closely in some modern editions . An ex- cellent example of this method of exhibiting the Elements of Euclid will be found in an edition in quarto , published at the Hague , in the French language , in 1762. Two per- sons are named ...
... less closely in some modern editions . An ex- cellent example of this method of exhibiting the Elements of Euclid will be found in an edition in quarto , published at the Hague , in the French language , in 1762. Two per- sons are named ...
Seite 6
... less than two right angles , these straight lines , being continually produced , shall at length meet on that side on which are the angles which are less than two right angles . I PROPOSITION 1. PROBLEM . To describe an equilateral ...
... less than two right angles , these straight lines , being continually produced , shall at length meet on that side on which are the angles which are less than two right angles . I PROPOSITION 1. PROBLEM . To describe an equilateral ...
Seite 9
... less . From the point A draw the straight line AD equal to C ; [ I. 2 . and from the centre A , at the distance AD , describe the circle DEF meeting AB at E. [ Postulate 3 . AE shall be equal to C. Because the point A is the centre of ...
... less . From the point A draw the straight line AD equal to C ; [ I. 2 . and from the centre A , at the distance AD , describe the circle DEF meeting AB at E. [ Postulate 3 . AE shall be equal to C. Because the point A is the centre of ...
Seite 12
... less , and join DC . Then , because in the triangles DBC , ACB , DB is equal to AC , and BC is common to both , B LI . 3 . [ Construction . the two sides DB , BC are equal to the two sides AC , CB , each to each ; and the angle DBC is ...
... less , and join DC . Then , because in the triangles DBC , ACB , DB is equal to AC , and BC is common to both , B LI . 3 . [ Construction . the two sides DB , BC are equal to the two sides AC , CB , each to each ; and the angle DBC is ...
Seite 17
... less to the greater ; which is impossible . [ Axiom 9 : Wherefore two straight lines cannot have a common segment . PROPOSITION 12. PROBLEM . To draw a straight line perpendicular to a given straight line of an unlimited length , from a ...
... less to the greater ; which is impossible . [ Axiom 9 : Wherefore two straight lines cannot have a common segment . PROPOSITION 12. PROBLEM . To draw a straight line perpendicular to a given straight line of an unlimited length , from a ...
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ABCD AC is equal angle ABC angle ACB angle BAC angle EDF angles equal Axiom base BC bisects the angle centre chord circle ABC circle described circumference Construction Corollary describe a circle diameter double draw a straight equal angles equal to F equiangular equilateral equimultiples Euclid Euclid's Elements exterior angle given circle given point given straight line gnomon Hypothesis inscribed intersect isosceles triangle less Let ABC magnitudes middle point multiple opposite angles opposite sides parallelogram perpendicular plane polygon produced proportionals PROPOSITION 13 Q.E.D. PROPOSITION quadrilateral radius rectangle contained rectilineal figure remaining angle rhombus right angles right-angled triangle segment shew shewn side BC square on AC straight line &c straight line drawn tangent THEOREM touches the circle triangle ABC triangle DEF twice the rectangle vertex Wherefore