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

### From inside the book

Results 6-10 of 88

Page 10

... two sides might be produced in either direction . Hence the no . of

constructions is 2 x 2 x 2 , or 8 . Exceptional case , when A is situated at the

vertex of an equilat . A on BC . 2. In fig . to Prop . 15 let EX , EY

AED respectively .

... two sides might be produced in either direction . Hence the no . of

constructions is 2 x 2 x 2 , or 8 . Exceptional case , when A is situated at the

vertex of an equilat . A on BC . 2. In fig . to Prop . 15 let EX , EY

**bisect**_ S BEC ,AED respectively .

Page 11

21 , suppose BD , CD

= sum of _ * DBA , BAD ; < CDF = sum of _ S DCA , CAD . Hence LBDC = the

angle at A together with half the sum of the base angles . 7. In A ABC let the

external ...

21 , suppose BD , CD

**bisect**base < 5 , and let Ad be produced to F ; then _ BDF= sum of _ * DBA , BAD ; < CDF = sum of _ S DCA , CAD . Hence LBDC = the

angle at A together with half the sum of the base angles . 7. In A ABC let the

external ...

Page 12

the required A ( 1. 37 , and 1. 4 ] . 5. The As are on equal bases and of equal

altitude . 6. In the fig . to 1. 34 let the diagonals intersect in E. Then , by Ex . 5 on p

.

**Bisect**AB in E , and through E draw EF perp . to AB meeting CD in F. Then AFB isthe required A ( 1. 37 , and 1. 4 ] . 5. The As are on equal bases and of equal

altitude . 6. In the fig . to 1. 34 let the diagonals intersect in E. Then , by Ex . 5 on p

.

Page 14

3. It is easy to see that AS ACX , BCY are equal in all respects [ 1. 4 ] . .. AX = BY .

Similarly BY = cz . 4. Let BD be the diagonal of sq . ABCD . Then DB = sum of sqq

. on DC , BC = twice the sq . on DC . 5 . AX

3. It is easy to see that AS ACX , BCY are equal in all respects [ 1. 4 ] . .. AX = BY .

Similarly BY = cz . 4. Let BD be the diagonal of sq . ABCD . Then DB = sum of sqq

. on DC , BC = twice the sq . on DC . 5 . AX

**bisects**BC ( I. 26 ] ; .. sq . on BC = 4 ... Page 16

The exceptional case is when o is at the intersection of the diagonals . 8. Let the

median AD

AS ABD , EDC are identically equal ( 1. 4 ] and AB = CE . Now AC + CE > AE ( 1.

The exceptional case is when o is at the intersection of the diagonals . 8. Let the

median AD

**bisect**BC ; produce AD to E making DE equal to AD , join EC . ThenAS ABD , EDC are identically equal ( 1. 4 ] and AB = CE . Now AC + CE > AE ( 1.

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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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