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 Archi- medes ; but in following nearly these illustrious models I have endeavoured to improve certain ...
... solutions . I have thought it proper , therefore , to adopt in this work the same method which we find in the writings of Euclid and Archi- medes ; but in following nearly these illustrious models I have endeavoured to improve certain ...
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 34
... 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 35
... 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 36
... 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 , des- cribe 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 , des- cribe 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 BAD 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 and parallel equiangular equilateral equivalent faces figure four right angles frustum Geom gles greater hence homologous sides hypothenuse inclination inscribed circle isosceles join less let fall line AC manner mean proportional measure the half meet multiplied number of sides oblique lines opposite parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism proposition pyramid S-ABC quadrilateral radii radius ratio rectangle regular polygon right angles right-angled triangle Scholium segment semicircumference side BC similar solid angle sphere spherical polygons spherical triangle square described straight line tangent THEOREM three plane angles triangle ABC triangular prism triangular pyramids vertex vertices whence