## The Elements of Euclid, Viz: The Errors, by which Theon, Or Others, Have Long Ago Vitiated These Books, are Corrected; and Some of Euclid's Demonstrations are Restored. Also the Book of Euclid's Data, in Like Manner Corrected. the first six books, together with the eleventh and twelfth |

### From inside the book

Results 1-5 of 97

Page 17

...

...

**BC**fhall be equal to the bafe EF ; and the triangle ABC to the triangle DEF ; and the other angles , to which the ...**fide**of the bafe fhall be equal . Let ABC be an Isofceles triangle , of which the**fide**AB is e- B qual Book I. qual to ... Page 18

...

...

**BC**is common to the two triangles BFC , CGB ; wherefore the triangles are equal , and their remaining angles , each ...**fide**of the bafe . Therefore the angles at the bafe , & c . Q. E. D. COROLLARY . Hence every equilateral triangle is ... Page 19

...

...

**fide**AB is alfo equal to the**fide**AC . For , if AB be not equal to AC , one of them is greater than the other : Let ...**BC**common to both , the two fides DB ,**BC**are equal to the two AC , CB , each to each ; and the angle DBC is equal to ... Page 20

...

...

**fide**of the bafe CD are equal to one another ; but the angle ECD is greater ...**fide**of the other needs no de- monftration . a B Therefore upon the fame bafe , and on the ...**BC BC**is equal to EF ; therefore**BC**coinciding with 20 THE ELEMENTS. Page 21

...

...

**BC**is equal to EF ; therefore**BC**coinciding with EF , BA and Book J. AC ...**fide**of it , there can be two triangles that have their fides which are ...**BC**coin- a 7.1 . cides with the bafe EF , the fides BA , AC cannot but coincide with ...### Other editions - View all

### Common terms and phrases

alfo alſo angle ABC angle BAC bafe baſe BC is equal BC is given becauſe the angle bifected Book XI cafe circle ABCD circumference cone confequently cylinder defcribed demonftrated diameter drawn equal angles equiangular equimultiples Euclid excefs faid fame manner fame multiple fame ratio fame reafon fecond fegment fhall fhewn fide BC fimilar firft firſt folid angle fome fore fphere fquare of AC ftraight line AB given angle given ftraight line given in fpecies given in magnitude given in pofition given magnitude given ratio gnomon greater join lefs likewife line BC oppofite parallel parallelepipeds parallelogram perpendicular polygon prifm propofition proportionals pyramid Q. E. D. PROP rectangle contained rectilineal figure right angles thefe THEOR theſe triangle ABC vertex wherefore

### Popular passages

Page 32 - 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 165 - D ; wherefore the remaining angle at C is equal to the remaining angle at F ; Therefore the triangle ABC is equiangular to the triangle DEF.

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 10 - When several angles are at one point B, any ' one of them is expressed by three letters, of which ' the letter that is at the vertex of the angle, that is, at ' the point in which the straight lines that contain the ' angle meet one another, is put between the other two ' letters, and one of these two is...

Page 55 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.

Page 32 - ... then shall the other sides be equal, each to each; and also the third angle of the one to the third angle of the other. Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, viz.

Page 45 - To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Page 211 - AB shall be at right angles to the plane CK. Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK ; take any point F in CE, from which draw FG in the plane DE at right D angles to CE ; and because AB is , perpendicular to the plane CK, therefore it is also perpendicular to every straight line in that plane meeting it (3.

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.

Page 304 - Thus, if B be the extremity of the line AB, or the common extremity of the two lines AB, KB, this extremity is called a point, and has no length : For if it have any, this length must either be part of the length of the line AB, or of the line KB.