Elements of Geometry |
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Page 36
... tangent and a chord , has for its measure the half of the arc AMDC , compre- hended between its sides . Demonstration . At the point of contact A draw the diameter AD ; the angle BAD is a right angle ( 110 ) , and has for its measure ...
... tangent and a chord , has for its measure the half of the arc AMDC , compre- hended between its sides . Demonstration . At the point of contact A draw the diameter AD ; the angle BAD is a right angle ( 110 ) , and has for its measure ...
Page 41
... tangent required . For , if we draw CB , the angle CBA inscribed in a semicircle is a right angle ( 128 ) ; therefore AB , being a perpendicular at the extremity of the radius CB , is a tangent . 152. Scholium . The point A being ...
... tangent required . For , if we draw CB , the angle CBA inscribed in a semicircle is a right angle ( 128 ) ; therefore AB , being a perpendicular at the extremity of the radius CB , is a tangent . 152. Scholium . The point A being ...
Page 42
... tangent . The same may be said of the sides BC , AC . 154. Scholium The three lines , which bisect the three an- gles of a triangle , meet in the same point . PROBLEM . 155. Upon a given straight line AB ( fig . 88 , 89 ) to describe a ...
... tangent . The same may be said of the sides BC , AC . 154. Scholium The three lines , which bisect the three an- gles of a triangle , meet in the same point . PROBLEM . 155. Upon a given straight line AB ( fig . 88 , 89 ) to describe a ...
Page 68
... tangent OA be drawn , and a secant OC , the tangent will be a mean proportional between the secant and the part without 2 the circle ; that is , OC : OA :: ' OA : OD , or , OA = OC × OD . Demonstration . By joining AD and AC , the ...
... tangent OA be drawn , and a secant OC , the tangent will be a mean proportional between the secant and the part without 2 the circle ; that is , OC : OA :: ' OA : OD , or , OA = OC × OD . Demonstration . By joining AD and AC , the ...
Page 73
... tangent ; and , if AC be produced till it meet the circumference in E , we shall have AE : AB :: AB : AD ( 228 ) , and hence AE — AB : AB :: AB — AD : AD ( IV ) . But , since the radius BC is half of AB , the diameter DE is equal to AB ...
... tangent ; and , if AC be produced till it meet the circumference in E , we shall have AE : AB :: AB : AD ( 228 ) , and hence AE — AB : AB :: AB — AD : AD ( IV ) . But , since the radius BC is half of AB , the diameter DE is equal to AB ...
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Common terms and phrases
ABC fig adjacent angles altitude angle ACB angle BAC base ABCD bisect centre chord circ circular sector circumference circumscribed common cone consequently convex surface Corollary cube cylinder Demonstration diagonals diameter draw drawn equal angles equiangular equilateral equivalent figure formed four right angles frustum GEOM given point gles greater hence homologous sides hypothenuse inclination intersection isosceles triangle JOHN CRERAR LIBRARY join less Let ABC let fall line AC mean proportional measure the half meet multiplied number of sides oblique lines opposite parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism produced proposition pyramid S-ABC radii radius ratio rectangle regular polygon right angles Scholium sector segment semicircle semicircumference side BC similar solid angle sphere spherical polygons spherical triangle square described straight line tangent THEOREM three angles triangle ABC triangular prism triangular pyramids vertex vertices whence