Euclidian GeometryMacmillan, 1874 - 349 sider |
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Resultat 1-5 av 42
Side 2
... might 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 through INTRODUCTION .
... might 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 through INTRODUCTION .
Side 3
Francis Cuthbertson. 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. P A Then ...
Francis Cuthbertson. 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. P A Then ...
Side 8
... .. L FHD is L EHD is = L'EDH ; LFDH ; ( I. 2 ) ( 1. 2 ) and · . FH is – FD , .. the whole EHF is the whole △ EDF , * and ..L BAC is = LEDF . ( ii ) If HD passes through one extremity of 8 STRAIGHT LINES , ANGLES AND TRIANGLES .
... .. L FHD is L EHD is = L'EDH ; LFDH ; ( I. 2 ) ( 1. 2 ) and · . FH is – FD , .. the whole EHF is the whole △ EDF , * and ..L BAC is = LEDF . ( ii ) If HD passes through one extremity of 8 STRAIGHT LINES , ANGLES AND TRIANGLES .
Side 9
Francis Cuthbertson. ( ii ) If HD passes through one extremity of EF as F. 4 A E · EH is = ED , .. LEHD is = LEDH ; :: and .. △ BAC is = LEDF . ( iii ) If HD does not meet EF . H ( I. 2 ) H · EH is = ED , .. LEHD is = △ EDH ; ( 1. 2 ) ...
Francis Cuthbertson. ( ii ) If HD passes through one extremity of EF as F. 4 A E · EH is = ED , .. LEHD is = LEDH ; :: and .. △ BAC is = LEDF . ( iii ) If HD does not meet EF . H ( I. 2 ) H · EH is = ED , .. LEHD is = △ EDH ; ( 1. 2 ) ...
Side 28
... . and is .. > EQF . Much more then is EFQ > LEQF ; .. EQ is > EF ; i.e. BC is 2nd . Let EQ pass through F EF . ( 1. 15 ) B A E Then it is evident that EQ is > EF . i.e. BC is > EF . F 3rd . Let EQ not meet DF . · A 28.
... . and is .. > EQF . Much more then is EFQ > LEQF ; .. EQ is > EF ; i.e. BC is 2nd . Let EQ pass through F EF . ( 1. 15 ) B A E Then it is evident that EQ is > EF . i.e. BC is > EF . F 3rd . Let EQ not meet DF . · A 28.
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