## Euclidian Geometry |

### From inside the book

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Page vi

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**meet**. This is of great consequence , as the leading idea in connection with straight lines called parallel is that suggested by the literal meaning of the word : thus the conception of a paral- lelogram is not that of a figure whose ... Page 9

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**meet**EF . E H · . · EH is = ED , and ·· FH is = FD , .. LEHD is = △ EDH ; ..L FHD is = △ FDH ; .. L EHF is = L EDF , ( 1. 2 ) ( I. 2 ) and ..L BAC is = LEDF . Thus in any case the L BAC is = L EDF , also BA , AC are respectively = ED ... Page 27

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**meet**the side AC in E. Then BA and AE are together greater than BE ; add to each EC ; ( I. 16 ) .. BA and AC are together greater than BE and EC . Again , DE and EC are together greater than DC ; add to each BD ; ( I. 16 ) .. BE and EC ... Page 28

... is .. > EQF . Much more then is 4 EFQ > LEQF ; ... EQ is > EF ; i.e. BC is > EF . 2nd . Let EQ pass through F ( I. 15 ) E Then it is evident that EQ is > EF . i.e. BC is > EF . D 3rd . Let EQ not

... is .. > EQF . Much more then is 4 EFQ > LEQF ; ... EQ is > EF ; i.e. BC is > EF . 2nd . Let EQ pass through F ( I. 15 ) E Then it is evident that EQ is > EF . i.e. BC is > EF . D 3rd . Let EQ not

**meet**DF . J B 28 INEQUALITIES . Page 29

Francis Cuthbertson (M.A.). 3rd . Let EQ not

Francis Cuthbertson (M.A.). 3rd . Let EQ not

**meet**DF . J B Then EQ , QD are together > EF , FD , ( 1. 17 ) and QD is FD ; = .. EQ is EF , i.e. BC is > EF . > PROPOSITION XIX . If two triangles have two sides of the one equal to two sides ...### Other editions - View all

### Common terms and phrases

Algebra base Cambridge centre chord circumference cloth Conic Sections Crown 8vo Describe a circle diagonals diameter divided draw a straight ELEMENTARY TREATISE English equiangular equilateral Euclid Extra fcap fcap GEOMETRY given angle given circle given point given straight line Grammar greater H Let Hence inscribed intersecting isosceles triangle Latin Let ABC line bisecting locus Mathematical meet opposite angles Owens College parallel parallelogram perimeter perpendicular plane polygon PROBLEM produced Professor proportional PROPOSITION ratio rect rectangle rectangle contained rectilineal figure regular polygon respectively rhombus right angles Schools Second Edition segment similar Similarly squares on AC straight line drawn straight line joining tangent THEOREM TRIGONOMETRY twice rectangle twice the squares vertex