## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh and Twelfth |

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Page 7

Because the point A is the centre of the circle BCD , AC is

Because the point A is the centre of the circle BCD , AC is

**equal*** to AB ; and because the point B is the * 15 Deficentre of the circle ACE , BC is**equal**to BA : but it has nition . been proved that CA is**equal**to AB ; therefore CA ... Page 8

1 Ax . that BC is

1 Ax . that BC is

**equal**to BG ; wherefore AL and BC are each of them**equal**to BG : and things that are**equal**to the same thing are equalt to one another ; therefore the straight line AL is**equal**to BC . Wherefore from the given point A ... Page 9

to which the

to which the

**equal**sides are opposite , shall be**equal**each to each , viz . the angle ABC to the angle DEF , and the angle ACB to DFE . For , if the triangle ABC be applied to DEF , so that the point A may be on D , and the straight ... Page 10

FA G IE * 3 Ax . the two triangles AFC , AGB ; therefore the base FC is

FA G IE * 3 Ax . the two triangles AFC , AGB ; therefore the base FC is

**equal*** to the base GB , and the triangle AFC to the triangle AGB ; and the remaining angles of the one are**equal*** to the remaining angles of the other , each to ... Page 12

Again , because CB is

Again , because CB is

**equal**t to DB , the angle BDC is**equal*** to the angle BCD ; but BDC has been proved to be greater than the same BCD ; which is impossible . The case in which the vertex of one triangle is upon a side of the other ...### What people are saying - Write a review

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The Elements of Euclid: Viz. The First Six Books, Together With the Eleventh ... Euclid Euclid No preview available - 2018 |

### Common terms and phrases

added altitude angle ABC angle BAC base Book centre circle circle ABCD circumference common cone contained cylinder definition demonstrated described diameter difference divided double draw drawn equal equal angles equiangular equimultiples Euclid excess figure fore four fourth given angle given in position given in species given magnitude given ratio given straight line greater Greek half join less likewise logarithm magnitude manner meet multiple opposite parallel parallelogram pass perpendicular plane prism produced PROP proportionals proposition proved pyramid radius reason rectangle rectilineal figure remaining right angles segment shewn sides similar sine solid sphere square square of AC taken THEOR third triangle ABC wherefore whole