The Elements of EuclidLongmans, Green and Company, 1866 - 376 páginas |
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Página 45
... shewn to be equilateral ; therefore it is a square ( Dɛf . 30 ) , and it is described upon the given straight line AB . Q. E. F. COR . Hence every parallelogram that has one right angle has all its angles right angles . PROP . XLVII ...
... shewn to be equilateral ; therefore it is a square ( Dɛf . 30 ) , and it is described upon the given straight line AB . Q. E. F. COR . Hence every parallelogram that has one right angle has all its angles right angles . PROP . XLVII ...
Página 50
... shewn that DE is equal to DG ; therefore DF is equal to DG . Now , the square on CD is equal to the squares on CF , DF , and also equal to the squares on CG , DG ; hence , as DF is equal to DG , CF must be equal to CG ; and , therefore ...
... shewn that DE is equal to DG ; therefore DF is equal to DG . Now , the square on CD is equal to the squares on CF , DF , and also equal to the squares on CG , DG ; hence , as DF is equal to DG , CF must be equal to CG ; and , therefore ...
Página 82
... shewn to equal twice the rectangle AB , AE together with the square on BD ; therefore the rectangle AB , AE is equal to the rectangle AC , AD . 6. C is a given point in a given straight line AB : It is required to find a point F in CB ...
... shewn to equal twice the rectangle AB , AE together with the square on BD ; therefore the rectangle AB , AE is equal to the rectangle AC , AD . 6. C is a given point in a given straight line AB : It is required to find a point F in CB ...
Página 92
... shewn that no other point but F is the centre ; that is , F has been found , the centre of the circle ABC . Q. E. F. COR . From this it is manifest that , if in a circle one straight line bisect another at right angles , the centre of ...
... shewn that no other point but F is the centre ; that is , F has been found , the centre of the circle ABC . Q. E. F. COR . From this it is manifest that , if in a circle one straight line bisect another at right angles , the centre of ...
Página 95
... shewn to be a right angle ; there- fore the angle FEA is equal to the angle FEB , the less to the greater — which is absurd : Therefore AC , BD do not bisect one another . Wherefore , If in a circle , & c . Q. E. D. PROP . V. THEOR . If ...
... shewn to be a right angle ; there- fore the angle FEA is equal to the angle FEB , the less to the greater — which is absurd : Therefore AC , BD do not bisect one another . Wherefore , If in a circle , & c . Q. E. D. PROP . V. THEOR . If ...
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