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

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Page 11

AB , each to each ; and they contain the angle FAG

AB , each to each ; and they contain the angle FAG

**common**to the two triangles AFC , AGB ; therefore the A base FC is equal * to the base GB , and the triangle AFC to the triangle AGB ; and the remaining angles of the one are В , equal ... Page 12

... and the base BC is

... and the base BC is

**common**to the two triangles BFC , CGB ; wherefore these triangles are equal * , and their remaining angles each to each , to which the equal sides are opposite : therefore the angle FBC is equal to the angle GCB ... Page 15

Because AD is equalt to AE , DA E | Const . and AF is

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. # Const , Because ACis equalf to CB , and CD

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

By help of this problem , it may be demonstrated , that two straight lines cannot have a

By help of this problem , it may be demonstrated , that two straight lines cannot have a

**common**segment . If it be possible , let the two straight lines ABC , ABD have the segment AB**common**to both of them .### 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