The Elements of Euclid for the Use of Schools and CollegesMacmillan, 1872 - 400 sider |
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Resultat 1-5 av 53
Side 7
... describe the circle BCD . [ Postulate 3 . From the centre B , at the distance BA , describe the circle ACE . [ Postulate 3 . From the point C , at which the circles cut one another , draw the straight lines CA and CB to the points A and ...
... describe the circle BCD . [ Postulate 3 . From the centre B , at the distance BA , describe the circle ACE . [ Postulate 3 . From the point C , at which the circles cut one another , draw the straight lines CA and CB to the points A and ...
Side 8
... describe the circle GKL , meeting DE at L. [ Post . 3 . AL shall be equal to BC . K B G Because the point B is the centre of the circle CGH , BC is equal to BG . [ Definition 15 . And because the point D is the centre of the circle GKL ...
... describe the circle GKL , meeting DE at L. [ Post . 3 . AL shall be equal to BC . K B G Because the point B is the centre of the circle CGH , BC is equal to BG . [ Definition 15 . And because the point D is the centre of the circle GKL ...
Side 9
... describe the circle DEF meeting AB at E. [ Postulate 3 . AE shall be equal to C. Because the point A is the centre of the circle DEF , AE is equal to AD . But C is equal to AD . [ Definition 15 . [ Construction . Therefore AE and Care ...
... describe the circle DEF meeting AB at E. [ Postulate 3 . AE shall be equal to C. Because the point A is the centre of the circle DEF , AE is equal to AD . But C is equal to AD . [ Definition 15 . [ Construction . Therefore AE and Care ...
Side 17
... describe the circle EGF , meeting AB at F and G. Bisect FG at H , and join CH . [ Postulate 3 . [ I. 10 . The straight line CH drawn from the given point C shall be perpendicular to the given straight line AB . Join CF , CG . Because FH ...
... describe the circle EGF , meeting AB at F and G. Bisect FG at H , and join CH . [ Postulate 3 . [ I. 10 . The straight line CH drawn from the given point C shall be perpendicular to the given straight line AB . Join CF , CG . Because FH ...
Side 25
... describe the circle DKL . [ Post . 3 . D -E H L A B- From the centre G , at the distance GH , describe the circle HLK , cutting the former circle at K. Join KF , KG . The triangle KFG shall have its sides equal to the three straight ...
... describe the circle DKL . [ Post . 3 . D -E H L A B- From the centre G , at the distance GH , describe the circle HLK , cutting the former circle at K. Join KF , KG . The triangle KFG shall have its sides equal to the three straight ...
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Vanlige uttrykk og setninger
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