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

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Page 9

Because the point B is the K centre of the circle CGH , BC is equal * to BG ; and because D H is the centre of the circle GKL , D DL is equal to DG ; and † DA ,

Because the point B is the K centre of the circle CGH , BC is equal * to BG ; and because D H is the centre of the circle GKL , D DL is equal to DG ; and † DA ,

**Const**. BI DB , parts of them , are equal ; therefore the remainder AL is F ... Page 11

Because AF is equal to + AG , and AB tofAC ,

Because AF is equal to + AG , and AB tofAC ,

**Const**. the two sides FA , AC are equal to the two GA , 6 Hyp . AB , each to each ; and they contain the angle FAG common to the two triangles AFC , AGB ; therefore the A base FC is equal ... Page 15

Because AD is equalt to AE , DA E |

Because AD is equalt to AE , DA E |

**Const**. and AF is common to the two , triangles DAF , EAF ; the two sides DA , AF , are equal to the B F two sides EA , AF , each to each ; and the base DF , is equalt to the base EF ; there- 4**Const**... Page 16

AB shall be cut into two equal parts in the point D. #

AB shall be cut into two equal parts in the point D. #

**Const**, Because ACis equalf to CB , and CD common to the two triangles с ACD , BCD ; the two sides XC , CD are equal to BC , CD , each to each ; and the angle ACD is †**Const**. equal ... Page 17

each to each : and the base DF is equalt to the +

each to each : and the base DF is equalt to the +

**Const**. base EF ; therefore the angle DCF is equal * to * 8 . 1 . the angle ECF : and they are adjacent angles . But when the adjacent angles which one straight line makes with another ...### 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

altitude angle ABC angle BAC base base BC BC is equal centre circle ABCD circumference common cone Const contained cylinder demonstrated described diameter divided double draw drawn equal angles equiangular equilateral equimultiples extremities fall figure fore four fourth given given straight line greater half inscribed join less Let ABC likewise magnitudes manner meet multiple opposite parallel parallelogram pass perpendicular plane polygon prism PROB produced PROP proportionals proved pyramid ratio reason rectangle contained rectilineal figure remaining angle right angles segment shown sides similar solid solid angle solid parallelopiped sphere square taken THEOR third touches triangle ABC vertex wherefore whole