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

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Page 15

Poft , CA , CB to the points A , B ;

Poft , CA , CB to the points A , B ;

**ABC**ball be an equilateral**triangle**. Because the point A is the centre of the circle BCD , AC is equal to AB ; and because the point B is the centre of the c . 15th Deo circle ACE , BC is equal to ... Page 17

For , if the

For , if the

**triangle ABC**be applied 10 DEF , so that the point A may be on D , and the straight line AB upon DE ; the point B shall coincide with the point E , because AB is equal to DE ; and AB coinciding with DE , AC shall coincide ... Page 18

G Book I. qual to AC , and let the straight lines AB , AC be produced to D and E , the angle

G Book I. qual to AC , and let the straight lines AB , AC be produced to D and E , the angle

**ABC**shall be equal to ... I. qualb to the base GB , and the**triangle**AFC to the**triangle**AGB ; and the remaining angles of the one are equal b ... Page 19

Let

Let

**ABC**be a**triangle**having the angle**ABC**equal to the Book 1 . angle ACB ; the fide AB is also equal to the side AC . ... I. qual to AC , the less , and join DC ; therefore , because in the**triangles**DBC , ACB , A DB is equal to AC ... Page 20

The case in which the ver . tex of one

The case in which the ver . tex of one

**triangle**is upon a side of the other , needs 110 de . monstration . ... Let**ABC**, DEF be two**triangles**having the two sides AB , AC equal to the two sides DE , DF , each to each , viz .### 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.