## A Key to the Exercises and Examples Contained in a Text-book of Euclid's Elements |

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Page 11

Hence LBDC = the angle at A together with half the sum of the base angles . 7. In

A ABC let the

at E. Then by Ex . 2 , page 29 each of _ SEBF , ECF is a rt . L. 1. LF is the suppt ...

Hence LBDC = the angle at A together with half the sum of the base angles . 7. In

A ABC let the

**external**bisectors of $ B and C meet at F , and the internal bisectorsat E. Then by Ex . 2 , page 29 each of _ SEBF , ECF is a rt . L. 1. LF is the suppt ...

Page 30

With the same lettering and construction as in Ex . 5 , let DE be produced both

ways indefinitely to X and Y. Then the required locus is the part of XY

the AODE . [ See Ex . 23 , p . 99. ] 7. Let AB be the rod of given length , and c the ...

With the same lettering and construction as in Ex . 5 , let DE be produced both

ways indefinitely to X and Y. Then the required locus is the part of XY

**external**tothe AODE . [ See Ex . 23 , p . 99. ] 7. Let AB be the rod of given length , and c the ...

Page 36

ABGH , BCEF be described

Be are clearly in one line , for the < 8 ABH , CBE are each half of a rt . angle . Also

as in 11. 9 , HE ? = 2HD . But by 11. 4 , HE ? = HB ? + BE + 2HB.BE , and HD ?

ABGH , BCEF be described

**externally**to the rectangle . Join HB , BE . Then HB ,Be are clearly in one line , for the < 8 ABH , CBE are each half of a rt . angle . Also

as in 11. 9 , HE ? = 2HD . But by 11. 4 , HE ? = HB ? + BE + 2HB.BE , and HD ?

Page 43

And as P is a fixed point , Q is determined . 41. This is the same as dividing a line

Draw the rect . ABDH , contained by THEOREMS AND EXAMPLES ON BOOK II .

And as P is a fixed point , Q is determined . 41. This is the same as dividing a line

**externally**in medial section . ( See Ex . 21 , p . 146 , and p . 139 , note . ] 42 .Draw the rect . ABDH , contained by THEOREMS AND EXAMPLES ON BOOK II .

Page 48

CAP – CB2 = 7 . But CA - CB = 1 . : . CA + CB , that is , AB + 2CB = 7 . .. CB = 3 . ::

. CE = CBP + BE ? = 32 + 42 = 25 . 1. CE ( the radius ) = 5 . 8. Let A , B be the

centres of the two circles , touching

CAP – CB2 = 7 . But CA - CB = 1 . : . CA + CB , that is , AB + 2CB = 7 . .. CB = 3 . ::

. CE = CBP + BE ? = 32 + 42 = 25 . 1. CE ( the radius ) = 5 . 8. Let A , B be the

centres of the two circles , touching

**externally**at C. Then ACB is a st . line [ 111.### What people are saying - Write a review

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A Key to the Exercises and Examples Contained in a Text-Book of Euclid's ... H S 1848-1934 Hall No preview available - 2018 |

A Key to the Exercises and Examples Contained in a Text-Book of Euclid's ... H. S. Hall No preview available - 2017 |

### Common terms and phrases

ABCD angles Assistant base bisector bisects BOOK Cambridge centre chord circle College common constant containing describe diam diameter distance double draw drawn Edited ELEMENTARY ENGLISH equal EXAMPLES Exercises external Fcap Fellow figure fixed four given given st greater GREEK half Hence HISTORY identically equal Illustrated inscribed intersect Introduction JOHN Join LATIN Let ABC locus Master Mathematics meet middle point Notes parl parm passes perp plane preparation Press produced Prof Professor proved radius ratio rect respectively revised School segment shewn sides similar Similarly solutions Take tangent touch Translated triangle twice University vols

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