Elements of Geometry and Trigonometry |
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Page 42
... fall exactly on the A curve line AFB , otherwise there would , in the one or the other , be points unequally dis- tant from the centre , which is contrary to the definition of a circle . PROPOSITION II . THEOREM . -Every chord is less ...
... fall exactly on the A curve line AFB , otherwise there would , in the one or the other , be points unequally dis- tant from the centre , which is contrary to the definition of a circle . PROPOSITION II . THEOREM . -Every chord is less ...
Page 44
... fall on G ; therefore , the chord AD is equal to the chord EG . Conversely , supposing again the radii AC , EO , to be equal , if the chord AD is equal to the chord EG , the arcs AMD , ENG will also be equal . For , if the radii CD , OG ...
... fall on G ; therefore , the chord AD is equal to the chord EG . Conversely , supposing again the radii AC , EO , to be equal , if the chord AD is equal to the chord EG , the arcs AMD , ENG will also be equal . For , if the radii CD , OG ...
Page 45
... , this perpendicular is the same as the one let fall from the centre on the same chord , since both of them pass through the centre and middle of the chord . PROPOSITION VII . THEOREM . Through three given points not BOOK III . 45.
... , this perpendicular is the same as the one let fall from the centre on the same chord , since both of them pass through the centre and middle of the chord . PROPOSITION VII . THEOREM . Through three given points not BOOK III . 45.
Page 46
... fall from the same point B , on the same straight line , which is impossible ( Book I. Prop . XIV . ) ; hence DE , FG , will always meet in some point O. And moreover , this point O , since it lies in the perpendicular DE , is equally ...
... fall from the same point B , on the same straight line , which is impossible ( Book I. Prop . XIV . ) ; hence DE , FG , will always meet in some point O. And moreover , this point O , since it lies in the perpendicular DE , is equally ...
Page 47
... fall CF perpendicular to this chord , and CI perpendicu- lar to AH . It is evident that CF is greater than CO , and CO than CI ( Book I. Prop . XV . ) ; therefore , CF is still greater than CI . But CF is equal to CG , because the ...
... fall CF perpendicular to this chord , and CI perpendicu- lar to AH . It is evident that CF is greater than CO , and CO than CI ( Book I. Prop . XV . ) ; therefore , CF is still greater than CI . But CF is equal to CG , because the ...
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Common terms and phrases
adjacent adjacent angles altitude angle ACB angle BAC ar.-comp base multiplied bisect Book centre chord circ circumference circumscribed common cone consequently convex surface cylinder diagonal diameter dicular distance draw drawn equal angles equally distant equation equiangular equivalent figure formed four right angles frustum given angle given line gles greater homologous sides hypothenuse inscribed circle inscribed polygon intersection less Let ABC let fall logarithm measured by half number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedron polygon ABCDE prism proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle S-ABC Scholium secant secant line segment side BC similar solid angle solid described sphere spherical polygon spherical triangle square described straight line tang tangent THEOREM triangle ABC triangular prism vertex