Elements of Geometry |
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Page v
... solutions . I have thought it proper , therefore , to adopt in this work the same method which we find in the writings of Euclid and Ar- chimedes ; but , in following nearly these illustrious models , I have endeavoured to improve ...
... solutions . I have thought it proper , therefore , to adopt in this work the same method which we find in the writings of Euclid and Ar- chimedes ; but , in following nearly these illustrious models , I have endeavoured to improve ...
Page 3
... solution . A Lemma is a subsidiary truth employed in the demonstration of a theorem , or in the solution of a problem . The common name of Proposition is given indifferently to theorems , problems , and lemmas . A Corollary is a ...
... solution . A Lemma is a subsidiary truth employed in the demonstration of a theorem , or in the solution of a problem . The common name of Proposition is given indifferently to theorems , problems , and lemmas . A Corollary is a ...
Page 36
... Solution . From the points A and B , as centres , and with a radius greater than the half of AB , describe two arcs cutting each other in D ; the point D will be equally distant from the points A and B ; find , in like manner , either ...
... Solution . From the points A and B , as centres , and with a radius greater than the half of AB , describe two arcs cutting each other in D ; the point D will be equally distant from the points A and B ; find , in like manner , either ...
Page 37
... Solution . Take the points B and C , at equal distances from A ; and from B and C , as centres , with a radius greater than BA , describe two arcs cutting each other in D ; draw AD , which will be the perpendicular required . For , the ...
... Solution . Take the points B and C , at equal distances from A ; and from B and C , as centres , with a radius greater than BA , describe two arcs cutting each other in D ; draw AD , which will be the perpendicular required . For , the ...
Page 38
... Solution 1. If it is proposed to bisect the arc AB ( fig . 74 ) , from the points A and B , as centres , with the same radius , de- scribe two arcs intersecting each other in D ; through the point D and the centre C draw CD , which will ...
... Solution 1. If it is proposed to bisect the arc AB ( fig . 74 ) , from the points A and B , as centres , with the same radius , de- scribe two arcs intersecting each other in D ; through the point D and the centre C draw CD , which will ...
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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 construction convex surface Corollary cube cylinder Demonstration diagonals diameter draw drawn equal angles equiangular equilateral equivalent faces 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 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