The Synoptical Euclid; Being the First Four Books of Euclid's Elements of Geometry, from the Edition of Dr. Robert Simson ... With ExercisesCharles Henry Law, 1854 - 120 σελίδες |
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Σελίδα 49
... parallel to CD or BE ; then ( Constr . ) and through 1 . and The rectangle AE is equal to the rectangles AD , CE , 2. AE is the rectangle contained by AB , BC , for it is contained by AB , BE , of which BE is equal ( Def . 30. ) to BC ...
... parallel to CD or BE ; then ( Constr . ) and through 1 . and The rectangle AE is equal to the rectangles AD , CE , 2. AE is the rectangle contained by AB , BC , for it is contained by AB , BE , of which BE is equal ( Def . 30. ) to BC ...
Σελίδα 51
... CD , is equal to the square of CB . A C D B K L H M E G F Upon CB describe ( I. 46. ) the square CEFB , join BE , and through D draw ( I. 31. ) DHG parallel to CE or BF ; and through H draw KLM parallel to CB or EF ; and also through 4 ...
... CD , is equal to the square of CB . A C D B K L H M E G F Upon CB describe ( I. 46. ) the square CEFB , join BE , and through D draw ( I. 31. ) DHG parallel to CE or BF ; and through H draw KLM parallel to CB or EF ; and also through 4 ...
Σελίδα 52
... CD ; therefore 5. The gnomon CMG , together with LG , is equal to the rectangle AD , DB , together with the square ... parallel to CE or DF ; and through H draw KLM parallel to AD or EF ; and also through A draw AK parallel to CL ...
... CD ; therefore 5. The gnomon CMG , together with LG , is equal to the rectangle AD , DB , together with the square ... parallel to CE or DF ; and through H draw KLM parallel to AD or EF ; and also through A draw AK parallel to CL ...
Σελίδα 53
... CD ; therefore 6. The rectangle AD , DB , together with the square of CB , is equal to the square of CD . Wherefore ... parallel to AD or BE , and through G draw HGK parallel to AB or DE . And because AG is equal ( I. 43. ) to GE ...
... CD ; therefore 6. The rectangle AD , DB , together with the square of CB , is equal to the square of CD . Wherefore ... parallel to AD or BE , and through G draw HGK parallel to AB or DE . And because AG is equal ( I. 43. ) to GE ...
Σελίδα 56
... CD B EUCLID'S ELEMENTS . From the point C draw ( I. 11. ) CE at right angles to AB , and make it equal ( I. 3. ) to AC or CB , and join EA , EB ; through D draw ( I. 31. ) DF parallel to CE ; and through F draw FG parallel to AB ; and ...
... CD B EUCLID'S ELEMENTS . From the point C draw ( I. 11. ) CE at right angles to AB , and make it equal ( I. 3. ) to AC or CB , and join EA , EB ; through D draw ( I. 31. ) DF parallel to CE ; and through F draw FG parallel to AB ; and ...
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AB is equal AC is equal adjacent angles angle ABC angle ACB angle AGH angle BAC angle BCD angle EDF angle equal base BC bisected circle ABC circumference diameter double draw equal angles equal Constr equal Hyp equal straight lines equal to BC equilateral and equiangular EUCLID'S ELEMENTS exterior angle given circle given point given rectilineal angle given straight line given triangle gnomon greater inscribed interior and opposite less Let ABC Let the straight likewise opposite angles parallel to CD parallelogram pentagon perpendicular point F produced Q.E.D. PROP rectangle AE rectangle contained remaining angle required to describe right angles semicircle side BC square of AC straight line AB straight line AC straight line drawn touches the circle triangle ABC twice the rectangle