The First Six Books: Together with the Eleventh and Twelfth |
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Page 15
... fore equilateral , and it is described ' upon the given straight line AB . Which was required to be done , PRO P. II . - PROB .
... fore equilateral , and it is described ' upon the given straight line AB . Which was required to be done , PRO P. II . - PROB .
Page 19
A fore , because in the triangles DBC , ACB , DB is equal to AC , and BC common to D both , the two sides DB , BC are equal to the two AC , CB , each to ...
A fore , because in the triangles DBC , ACB , DB is equal to AC , and BC common to D both , the two sides DB , BC are equal to the two AC , CB , each to ...
Page 21
Axo fore if two triangles , & c . Q. E. D. PRO P. IX . PRO B. Ti . O bise & t a given rectilineal angle , that is , to divide it into two equal angles .
Axo fore if two triangles , & c . Q. E. D. PRO P. IX . PRO B. Ti . O bise & t a given rectilineal angle , that is , to divide it into two equal angles .
Page 24
Ax . fore the angles CBE , EBD are equal c to the three angles CBA , ABE , EBD . Again , because the angle DBA is equal to the two angles DBE , EBA , add to ...
Ax . fore the angles CBE , EBD are equal c to the three angles CBA , ABE , EBD . Again , because the angle DBA is equal to the two angles DBE , EBA , add to ...
Page 28
ABC would be less than the angle ACB ; but it is not ; there fore the side AC is not less than AB ; and it has been shewn that it is not equal to AB ...
ABC would be less than the angle ACB ; but it is not ; there fore the side AC is not less than AB ; and it has been shewn that it is not equal to AB ...
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added alſo altitude angle ABC angle BAC baſe becauſe Book caſe centre circle circle ABCD circumference common compounded cone cylinder definition demonſtrated deſcribed diameter divided double draw drawn equal equal angles equiangular equimultiples exceſs fame fides firſt folid fore four fourth given angle given in poſition given in ſpecies given magnitude given ratio given ſtraight line greater Greek half join leſs likewiſe magnitude manner meet muſt oppoſite parallel parallelogram perpendicular plane priſms produced prop proportionals propoſition pyramid reaſon rectangle rectangle contained remaining right angles ſame ſecond ſegment ſhall ſides ſimilar ſolid ſphere ſquare ſquare of AC Take taken theſe third triangle ABC wherefore whole
Popular passages
Page 462 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; and each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds ; and these into thirds, &c.
Page 160 - 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 81 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...
Page 105 - DEF are likewise equal (13. i.) to two right angles ; therefore the angles AKB, AMB are equal to the angles DEG, DEF, of which AKB is equal to DEG ; wherefore the remaining angle AMB is equal to the remaining angle DEF.
Page 157 - AC the same multiple of AD, that AB is of the part which is to be cut off from it : join BC, and draw DE parallel to it : then AE is the part required to be cut off.
Page 10 - 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.
Page 62 - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.
Page 112 - To describe an equilateral and equiangular pentagon about a given circle. • Let ABCDE be the given circle; it is required to describe an equilateral and equiangular pentagon about the circle ABCDE. Let the angles of a pentagon, inscribed in the circle...
Page 190 - 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...
Page 38 - F, which is the common vertex of the triangles ; that is, together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.