The Synoptical Euclid; Being the First Four Books of Euclid's Elements of Geometry from the Edition of Dr. Robert Simson ... With Exercises. By S. A. Good ... Second Edition |
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Página 19
From this it is manifest , that , if two straight lines cut one another , the angles they make at the point where ... it to F , and make EF equal to BE ; join also FC , and produce AC to G. Because AE is equal to EC , and BE to EF ; 1 .
From this it is manifest , that , if two straight lines cut one another , the angles they make at the point where ... it to F , and make EF equal to BE ; join also FC , and produce AC to G. Because AE is equal to EC , and BE to EF ; 1 .
Página 23
Take a straight line DE terminated at the point D , but unlimited towards E , and make ( I. 3. ) DF equal to 1 , FG equal to B , and GH equal to C ; and from the centre F , at a distance FD , describe ( Post . 3. ) ...
Take a straight line DE terminated at the point D , but unlimited towards E , and make ( I. 3. ) DF equal to 1 , FG equal to B , and GH equal to C ; and from the centre F , at a distance FD , describe ( Post . 3. ) ...
Página 28
Α E B G с F For , if it be not parallel , AB and CD being produced , shall meet either towards B , D , or towards 4 , C ; let them be produced and meet towards B , D , in the point G ; therefore 1 . GEF is a triangle , and its exterior ...
Α E B G с F For , if it be not parallel , AB and CD being produced , shall meet either towards B , D , or towards 4 , C ; let them be produced and meet towards B , D , in the point G ; therefore 1 . GEF is a triangle , and its exterior ...
Página 31
In BC take any point D , and join 4D ; and at the point 4 , in the straight line AD , make ( I. 23. ) the angle DAE equal to the angle ADC ' ; and produce the straight line FA to F ; EF ' is parallel to BC .
In BC take any point D , and join 4D ; and at the point 4 , in the straight line AD , make ( I. 23. ) the angle DAE equal to the angle ADC ' ; and produce the straight line FA to F ; EF ' is parallel to BC .
Página 32
E A For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides , by drawing straight lines from a point F within the figure to each of its angles . And by the preceding proposition , ali the angles of ...
E A For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides , by drawing straight lines from a point F within the figure to each of its angles . And by the preceding proposition , ali the angles of ...
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AB is equal ABC is equal ABCD angle ABC angle ACB angle AGH angle BAC angle BCD angle equal base BC bisected centre circle ABC circumference coincide common demonstrated describe diameter distance divided double draw equal angles exterior angle extremity fall figure four given circle given point given straight line gnomon greater impossible inscribed join less Let ABC Let the straight likewise manner meet opposite angles parallel parallelogram pass pentagon perpendicular point F PROBLEM produced Q.E.D. PROP reason rectangle contained rectilineal figure remaining angle right angles segment semicircle shown side BC sides square of AC straight line AC Take touches the circle triangle ABC twice the rectangle wherefore whole