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

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ROM the

ROM the

**greater**of two given ftraight lines to cut off a part equal to the less . Let AB and C be the two gi ven ftraight lines , whereof AB is the**greater**. It is required to cut off from AB , the**greater**, a part equal to C the lefs . Page 19

For , if AB be not equal to AC , one of them is

For , if AB be not equal to AC , one of them is

**greater**than a the other : Let AB be the**greater**, and from it cut off DB e a 3. I. qual to AC , the lefs , and join DC ; there- fore , because in the triangles DBC , ACB , DB is equal to ... Page 20

Again , becaufe CB is equal A to DB , the angle BDC is equal to the angle BCD ; but BDC has been proved to be

Again , becaufe CB is equal A to DB , the angle BDC is equal to the angle BCD ; but BDC has been proved to be

**greater**than the fame BCD ; which is impoffible . The cafe in which the ver tex of one triangle is upon a fide of the other ... Page 26

THEOR . one fide of a triangle be produced , the exterior angle is

THEOR . one fide of a triangle be produced , the exterior angle is

**greater**than either of the interior opposite angles . Let ABC be a triangle , and let its fide BC be produced to D , the exterior angle ACD is**greater**than either of the ... Page 27

thefe add the angle ACB ; therefore the angles ACD , ACB are Book I.

thefe add the angle ACB ; therefore the angles ACD , ACB are Book I.

**greater**than the angles ABC , ACB ; but ACD , ACB are to- gether equal b to two right angles ; therefore the angles ABC b . 13. 1 . BCA are less than two right angles ...### What people are saying - Write a review

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### Common terms and phrases

added alfo alſo altitude angle ABC angle BAC bafe baſe becauſe bifected Book Book XI centre circle circle ABCD circumference common cone cylinder defcribed definition demonftrated diameter divided double draw drawn equal equal angles equiangular equimultiples excefs fame fame multiple fecond fegment fhall fides fimilar firft folid folid angle fore four fourth fquare fquare of AC ftraight line given angle given in fpecies given in pofition given magnitude given ratio greater Greek half join lefs magnitude meet oppofite parallel parallelogram perpendicular plane prifms produced PROP propofition proportionals pyramid rectangle rectangle contained remaining right angles Take taken thefe THEOR theſe third triangle ABC wherefore whole

### Popular passages

Page 474 - 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.