The Elements of EuclidLongmans, Green and Company, 1866 - 376 páginas |
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Resultados 1-5 de 100
Página 9
... join FC GB . Then , because AF is equal to AG , and B F A AB to AC ( Hyp . ) , the two sides FA , AC are equal to the two GA , AB , each to each , and they contain the angle FAG common to the two triangles AFC , AGB - there- fore ( 1. 4 ) ...
... join FC GB . Then , because AF is equal to AG , and B F A AB to AC ( Hyp . ) , the two sides FA , AC are equal to the two GA , AB , each to each , and they contain the angle FAG common to the two triangles AFC , AGB - there- fore ( 1. 4 ) ...
Página 11
... join DC : Then , because in the triangles DBC , ACB , DB is equal to AC , and BC common to both triangles , the two sides DB , BC are equal to the two AC , CB , each to each , and the angle DBC is equal to the angle ACB ( Hyp ...
... join DC : Then , because in the triangles DBC , ACB , DB is equal to AC , and BC common to both triangles , the two sides DB , BC are equal to the two AC , CB , each to each , and the angle DBC is equal to the angle ACB ( Hyp ...
Página 13
... join DE , and upon DE , on the side opposite to that on which A lies , describe an equilateral triangle DEF , and join AF : the straight line AF bisects the angle BAC . D BFC Because AD is equal to AE , and AF is common to the two ...
... join DE , and upon DE , on the side opposite to that on which A lies , describe an equilateral triangle DEF , and join AF : the straight line AF bisects the angle BAC . D BFC Because AD is equal to AE , and AF is common to the two ...
Página 14
... DE describe the equi- lateral triangle DFE , and join CF : the straight line CF is drawn from the given point C at right angles to the given straight line AB . AD CE Because DC is equal to CE , and CF common 14 EUCLID'S ELEMENTS .
... DE describe the equi- lateral triangle DFE , and join CF : the straight line CF is drawn from the given point C at right angles to the given straight line AB . AD CE Because DC is equal to CE , and CF common 14 EUCLID'S ELEMENTS .
Página 19
... join BE , and produce it to F , making EF equal to BE ( 1. 3 ) ; and join FC . B A KE C D Then , because AE is equal to EC , and BE to EF , the two sides , AE , EB , are equal to the two , CE , EF , each to each , and the angle AEB is ...
... join BE , and produce it to F , making EF equal to BE ( 1. 3 ) ; and join FC . B A KE C D Then , because AE is equal to EC , and BE to EF , the two sides , AE , EB , are equal to the two , CE , EF , each to each , and the angle AEB is ...
Términos y frases comunes
AB² ABCD AC² AD² adjacent angles angle ABC angle ACB angle BAC angle equal base BC BC is equal centre chord circle ABC circumference diameter double draw equal angles equal to F equiangular equilateral triangle equimultiples exterior angle fore given circle given line given point given rectilineal given straight line gnomon hypotenuse inscribed intersection isosceles triangle join less Let ABC lines be drawn lines drawn meet multiple opposite angles parallel to BC parallelogram perpendicular plane polygon PROB produced proportionals Q. E. D. PROP rectangle contained rectilineal figure remaining angle right angles segment semicircle shew shewn square on AC tangents THEOR touches the circle triangle ABC twice the rectangle vertex Wherefore