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

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Page 8

AB to DE , and AC to DF ; and the

AB to DE , and AC to DF ; and the

**angle**B**BAC**equal to the**angle**EDF : the base BC shall be equal to the base EF ; and the triangle ABC to the triangle DEF ; and the other**angles**А CE to which the equal sides are opposite , shall be 8 ... Page 9

DE , AC shall coincide with DF , because the

DE , AC shall coincide with DF , because the

**angle BAC**is equal t to the angle EDF ; wherefore also the + Hyp . point C shall coincide with the point F , because the straight line AC + is equal to DF : but the point B was + Hyp . proved ... Page 12

The

The

**angle BAC**shall be equal to the angle EDF . For , if the triangle ABC be applied to DEF , so that the point B may be on E , and the straight line BC upon EF ; the point C shall also coincide with the point F , because BC is equal ... Page 13

A To bisect a given rectilineal

A To bisect a given rectilineal

**angle**, that is , to divide it into two equal**angles**. Let**BAC**be the given rectilineal**angle**; it is required to bisect it . Take any point D in AB , and from AC cut * off AE 3. Page 17

Again , because the straight line DE makes with AB the

Again , because the straight line DE makes with AB the

**angles**AED , DEB , these also are together equal to two right ... be produced to D : the exterior**angle**ACD shall be greater than either of the interior opposite**angles**CBA ,**BAC**.### 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