Euclidian Geometry |
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Resultat 1-5 av 36
Side 2
... be turned about its extremity A , towards the side on which BH is , so as to cut BH ; and thus two straight lines would enclose a space , which is impossible . • Hence it follows that If two straight lines pass 2 . INTRODUCTIO UCTION .
... be turned about its extremity A , towards the side on which BH is , so as to cut BH ; and thus two straight lines would enclose a space , which is impossible . • Hence it follows that If two straight lines pass 2 . INTRODUCTIO UCTION .
Side 3
Francis Cuthbertson (M.A.). • Hence it follows that If two straight lines pass through the same point they will coincide entirely or cut one another . For if not , if possible let them fall otherwise as AOB , POQ having a common point O ...
Francis Cuthbertson (M.A.). • Hence it follows that If two straight lines pass through the same point they will coincide entirely or cut one another . For if not , if possible let them fall otherwise as AOB , POQ having a common point O ...
Side 8
... ED , : . LEHD is = LEDH ; ( 1. 2 ) and FH is = FD , .. the whole .. EHF is the whole △ EDF , L FHD is = L FDH ; ( I. 2 ) and ... BAC is = LEDF . ( ii ) If HD passes through one extremity of 8 STRAIGHT LINES , ANGLES AND TRIANGLES .
... ED , : . LEHD is = LEDH ; ( 1. 2 ) and FH is = FD , .. the whole .. EHF is the whole △ EDF , L FHD is = L FDH ; ( I. 2 ) and ... BAC is = LEDF . ( ii ) If HD passes through one extremity of 8 STRAIGHT LINES , ANGLES AND TRIANGLES .
Side 9
Francis Cuthbertson (M.A.). ( ii ) If HD passes through one extremity of EF as F. A E H EH is = ED , L EHD is = L EDH ; ( 1. 2 ) ... and .. BAC is = LEDF . ( iii ) If HD does not meet EF . E H · . · EH is = ED , and ·· FH is = FD ...
Francis Cuthbertson (M.A.). ( ii ) If HD passes through one extremity of EF as F. A E H EH is = ED , L EHD is = L EDH ; ( 1. 2 ) ... and .. BAC is = LEDF . ( iii ) If HD does not meet EF . E H · . · EH is = ED , and ·· FH is = FD ...
Side 28
... > 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 .
... > 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 .
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