The Elements of EuclidJohnson and Warner, 1811 - 518 páginas |
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Página 15
... straight lines b CA , CB to the points A , B ; ABC shall be an equilateral triangle . D C Book I. a 3. Postu- late . A B E b 2. Post . Because the point A is the centre of the circle BCD , AC is equal to AB ; and because the point B is ...
... straight lines b CA , CB to the points A , B ; ABC shall be an equilateral triangle . D C Book I. a 3. Postu- late . A B E b 2. Post . Because the point A is the centre of the circle BCD , AC is equal to AB ; and because the point B is ...
Página 16
... straight line AL is equal to BC . Wherefore from the given point A a straight line AL has been drawn equal to the ... AC equal to the two sides DE , DF , each to each , viz . AB to DE , and AC to DF ; A 16 THE ELEMENTS.
... straight line AL is equal to BC . Wherefore from the given point A a straight line AL has been drawn equal to the ... AC equal to the two sides DE , DF , each to each , viz . AB to DE , and AC to DF ; A 16 THE ELEMENTS.
Página 17
Euclid Robert Simson. AB to DE , and AC to DF ; A and the angle BAC equal to ... line AB upon DE ; the point B shall coincide with the point E , because AB ... straight line AC is equal to DF : but the point B coincides with the point E ...
Euclid Robert Simson. AB to DE , and AC to DF ; A and the angle BAC equal to ... line AB upon DE ; the point B shall coincide with the point E , because AB ... straight line AC is equal to DF : but the point B coincides with the point E ...
Página 21
... AC shall coincide with ED and DF ; for , if the base BC coin- cides with the ... straight line AF bisects the angle BAC . sides EA , AF , Because AD is equal ... straight line AF , he base DF is ; therefore thè to the angle B D E Co & L ...
... AC shall coincide with ED and DF ; for , if the base BC coin- cides with the ... straight line AF bisects the angle BAC . sides EA , AF , Because AD is equal ... straight line AF , he base DF is ; therefore thè to the angle B D E Co & L ...
Página 22
... AC , CD are equal to BC , CD , each to each ; and the angle AGD is equal to the angle BCD ; there fore the base AD is equal to the base DB , and the straight line AB is divided into two equal parts in the point D. Which was to be done ...
... AC , CD are equal to BC , CD , each to each ; and the angle AGD is equal to the angle BCD ; there fore the base AD is equal to the base DB , and the straight line AB is divided into two equal parts in the point D. Which was to be done ...
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Términos y frases comunes
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