The Elements of Euclid |
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Page 37
... draw CE parallel to the straight line AB ; and be- cause AB is parallel to CE and AC meets them , the alternate ... drawing straight lines from a point F within the figure to each of its angles . And , by the preceding E D A B Book I ...
... draw CE parallel to the straight line AB ; and be- cause AB is parallel to CE and AC meets them , the alternate ... drawing straight lines from a point F within the figure to each of its angles . And , by the preceding E D A B Book I ...
Page 41
... draw a BE parallel to CA ; and through C draw CF pa- rallel to BD : therefore each of the figures EBCA , DBCF B A. D C F a 31. 1 . is a parallelogram ; and EBCA is equal to DBCF , because b 35. 1 . they are upon the same base BC , and ...
... draw a BE parallel to CA ; and through C draw CF pa- rallel to BD : therefore each of the figures EBCA , DBCF B A. D C F a 31. 1 . is a parallelogram ; and EBCA is equal to DBCF , because b 35. 1 . they are upon the same base BC , and ...
Page 42
... draw BG parallel to CA , and through F draw FH parallel to ED : then each of the figures GBCA , DEFH is a parallel- ogram ; and they are equal b to one another , because they are upon equal bases BC , EF , and be- tween the same pa ...
... draw BG parallel to CA , and through F draw FH parallel to ED : then each of the figures GBCA , DEFH is a parallel- ogram ; and they are equal b to one another , because they are upon equal bases BC , EF , and be- tween the same pa ...
Page 44
... draw AG parallel to EC , and through C draw CG parallel to EF : therefore FECG is a parallelo- gram : and because BE is equal to EC , the triangle ABE is like- wise equal d to the triangle AEC , since they are upon equal bases BE , EC ...
... draw AG parallel to EC , and through C draw CG parallel to EF : therefore FECG is a parallelo- gram : and because BE is equal to EC , the triangle ABE is like- wise equal d to the triangle AEC , since they are upon equal bases BE , EC ...
Page 46
... draw ↳ AH pa- rallel to BG or EF , and join HB . Then , because the straight line HF falls upon the parallels AH , EF , the angles AHF , HFE are together equal to two right angles ; wherefore the angles BHF , HFE are less than two ...
... draw ↳ AH pa- rallel to BG or EF , and join HB . Then , because the straight line HF falls upon the parallels AH , EF , the angles AHF , HFE are together equal to two right angles ; wherefore the angles BHF , HFE are less than two ...
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Common terms and phrases
altitude angle ABC angle BAC base BC BC is equal BC is given bisected Book XI centre circle ABCD circumference cone cylinder demonstrated described diameter draw drawn equal angles equiangular equimultiples Euclid excess fore given angle given in magnitude given in position given in species given magnitude given ratio given straight line gnomon greater join less Let ABC meet multiple parallel parallelogram parallelogram AC perpendicular point F polygon prism proportionals proposition pyramid Q. E. D. PROP radius ratio of AE rectangle CB rectangle contained rectilineal figure remaining angle right angles segment sides BA similar solid angle solid parallelopiped square of AC straight line AB straight line BC tangent THEOR third triangle ABC triplicate ratio vertex wherefore