The Elements of Euclid; viz. the first six books, together with the eleventh and twelfth. Also the book of Euclid's Data. By R. Simson. To which is added, A treatise on the construction of the trigonometrical canon [by J. Christison] and A concise account of logarithms [by A. Robertson].1834 |
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Page 12
... bisected by the straight line AF . Which was to be done . PROPOSITION X. PROB . To bisect a given finite straight line ; that is , to divide it into two equal parts . Let AB be the given straight line ; it is required to divide it into ...
... bisected by the straight line AF . Which was to be done . PROPOSITION X. PROB . To bisect a given finite straight line ; that is , to divide it into two equal parts . Let AB be the given straight line ; it is required to divide it into ...
Page 14
... bisect * FG in H , and join CH : the straight line CH , drawn from the given point C , shall be perpendicular to the given straight line AB . Join CF , CG and because FH is equal † to HG , and HC common to the two triangles FHC , GHC ...
... bisect * FG in H , and join CH : the straight line CH , drawn from the given point C , shall be perpendicular to the given straight line AB . Join CF , CG and because FH is equal † to HG , and HC common to the two triangles FHC , GHC ...
Page 16
... Bisect * AC in E ; join BE and produce it to F , and make EF equal to BE , and join FC . Because AE is equal † to EC ... bisected , and AC be produced to G , it may be demonstrated , that the angle BCG , that is 16 EUCLID'S ELEMENTS .
... Bisect * AC in E ; join BE and produce it to F , and make EF equal to BE , and join FC . Because AE is equal † to EC ... bisected , and AC be produced to G , it may be demonstrated , that the angle BCG , that is 16 EUCLID'S ELEMENTS .
Page 28
... bisect it . * C B Because AB is parallel to CD , and BC meets them , the alternate angles ABC , BCD are equal to one another : and because AC is parallel to BD , and BC meets them , the alternate angles ACB , CBD are equal to one ...
... bisect it . * C B Because AB is parallel to CD , and BC meets them , the alternate angles ABC , BCD are equal to one another : and because AC is parallel to BD , and BC meets them , the alternate angles ACB , CBD are equal to one ...
Page 33
... Bisect BC in E , join AE , and at the point E , in the straight line EC , make * the angle CEF equal to D ; and through A , draw * AFG parallel to EC , and through C , draw CG parallel to EF ; therefore FECG is a ↑ parallelogram . And ...
... Bisect BC in E , join AE , and at the point E , in the straight line EC , make * the angle CEF equal to D ; and through A , draw * AFG parallel to EC , and through C , draw CG parallel to EF ; therefore FECG is a ↑ parallelogram . And ...
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Common terms and phrases
ABC is given AC is equal altitude angle ABC angle BAC base BC bisected centre circle ABCD circle EFGH circumference common logarithm cone Constr cylinder demonstrated described diameter draw drawn equal angles equiangular equimultiples Euclid ex æquali excess fore given angle given in magnitude given in position given in species given magnitude given ratio given straight line gnomon greater join less Let ABC logarithm multiple parallel parallelogram perpendicular point F polygon prism Prop proportionals Q. E. D. PROPOSITION radius ratio of AE rectangle CB rectangle contained rectilineal figure remaining angle right angles segment shewn sides BA similar sine solid angle solid parallelopiped square of AC straight line AB straight line BC tangent THEOR.-If tiple triangle ABC vertex wherefore
Popular passages
Page 32 - 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...
Page 138 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Page 39 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.
Page 22 - If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another...
Page 41 - If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together •with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced.
Page 5 - If two triangles have two sides of the one equal to two sides of the other, each to each, but the...
Page 38 - IF a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square of the whole line. Let the straight line AB be divided...
Page 262 - Again ; the mathematical postulate, that " things which are equal to the same are equal to one another," is similar to the form of the syllogism in logic, which unites things agreeing in the middle term.
Page 89 - PBOR. —To describe an isosceles triangle, having each of the angles at the base, double of the third angle. Take any straight...
Page 165 - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.