## The First Six Books: Together with the Eleventh and Twelfth |

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Page 1

ALSO THE BOOK E U C L I D'S D AT A , In like

ALSO THE BOOK E U C L I D'S D AT A , In like

**manner**corrected . BY ROBERT SIMSON , M. D. meritus Professor of Mathematics in the University of Glasgow . To this SIXTH EDITION are also annexed , LEMENTS of PLAIN and SPHERICAL ... Page 23

E to the angle EBA ; in th fame

E to the angle EBA ; in th fame

**manner**, because ABD is a 10. Def . straight line , the angle DBE is equal to the angle EBA ; wherefore the angle DBE is equal to the angle CBE , the less to the D greater ; wbich is impossible ... Page 25

And , in like

And , in like

**manner**, it may be demonstrated , that no other can be in the same straight line with it but BD , which therefore is in the same Itraight line with CB . Wherefore , if at a point , & c . Q. E. D. PRO P. XV . THEOR . Page 26

... wherefore the angle BAE is equal to the angle ECF ; but the angle ECD is greater than the angle ECF ; therefore the angle ACD is greater than BAE : In the fame

... wherefore the angle BAE is equal to the angle ECF ; but the angle ECD is greater than the angle ECF ; therefore the angle ACD is greater than BAE : In the fame

**manner**, if the side BC be bisected , it may be demonstrated d . 15. Page 28

In the same

In the same

**manner**it may be demonstrated , that the ldes AB , BC are greater than CA , and BC , CA greater than AB . Therefore any two sides , & c . Q. E. D. C C19 1 . PRO P. XXI . THEO R. See N. Il ; from the ends of the side of a ...### What people are saying - Write a review

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### Common terms and phrases

added alſo altitude angle ABC angle BAC baſe becauſe Book caſe centre circle circle ABCD circumference common cone cylinder definition demonſtrated deſcribed diameter divided double draw drawn equal equal angles equiangular equimultiples exceſs fame fides figure 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 priſm produced prop proportionals propoſition pyramid reaſon rectangle rectangle contained rectilineal 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 483 - 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 170 - 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 167 - 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 200 - 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.