Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle and the Geometry of Solids ; to which are Added Elements of Plane and Spherical Trigonometry |
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Page 77
... perpendicular to AB , AE shall fall without the circle . In AE take any point F , join DF , and let DF meet the circle in C. Because DAF is a right angle , it is greater than the angle AFD ( 32. 1. ) ; but the greater angle of any ...
... perpendicular to AB , AE shall fall without the circle . In AE take any point F , join DF , and let DF meet the circle in C. Because DAF is a right angle , it is greater than the angle AFD ( 32. 1. ) ; but the greater angle of any ...
Page 79
... perpendicular to the line touching the circle . Let the straight line DE touch the circle ABC in the point C ; take the centre F , and draw the straight line FC : FC is perpendicular to DE . For , if it be not , from the point F draw ...
... perpendicular to the line touching the circle . Let the straight line DE touch the circle ABC in the point C ; take the centre F , and draw the straight line FC : FC is perpendicular to DE . For , if it be not , from the point F draw ...
Page 90
... perpendicular to AC ; there- fore AG is equal ( 3.3 . ) to GC ; wherefore F H AE.EC + ( 5.2 . ) EG2 AG2 , and adding A C G GF2 to both , AE.EC + EG2 + GF = AG2 + GF2 . Now EG2 + GF2 = EF2 , and B AG2 + GF2 = AF2 ; therefore AE.EC + EF2 ...
... perpendicular to AC ; there- fore AG is equal ( 3.3 . ) to GC ; wherefore F H AE.EC + ( 5.2 . ) EG2 AG2 , and adding A C G GF2 to both , AE.EC + EG2 + GF = AG2 + GF2 . Now EG2 + GF2 = EF2 , and B AG2 + GF2 = AF2 ; therefore AE.EC + EF2 ...
Page 91
... perpendicular ( 12 . 1. ) to AC , and join EB , EC , ED : and be- cause the straight line EF , which passes through the centre , cuts the straight line AC , which does not pass through the cen- tre , at right angles , it likewise ...
... perpendicular ( 12 . 1. ) to AC , and join EB , EC , ED : and be- cause the straight line EF , which passes through the centre , cuts the straight line AC , which does not pass through the cen- tre , at right angles , it likewise ...
Page 102
... perpendicular ( 18.3 . ) to KL ; therefore each of the angles at C is a right angle : for the same reason , the angles at the points B , D are right angles ; and because FCK is a right angle , the square of FK is equal ( 47. 1. ) to the ...
... perpendicular ( 18.3 . ) to KL ; therefore each of the angles at C is a right angle : for the same reason , the angles at the points B , D are right angles ; and because FCK is a right angle , the square of FK is equal ( 47. 1. ) to the ...
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ABC is equal ABCD altitude angle ABC angle ACB angle BAC angle EDF arch AC base BC bisected centre circle ABC circumference cosine cylinder definition demonstrated diameter draw equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given straight line greater hypotenuse inscribed join less Let ABC Let the straight line BC magnitudes MDCCCXX meet multiple opposite angle parallel parallelepipeds parallelogram perpendicular polygon prism PROB produced proportionals proposition Q. E. D. COR Q. E. D. PROP radius ratio rectangle contained rectilineal figure remaining angle segment semicircle shewn side BC sine solid angle solid parallelepipeds spherical angle spherical triangle straight line AC THEOR third touches the circle triangle ABC triangle DEF wherefore