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, Edlemnts of Plane and Spherical TrigonometryW.E. Dean, 1836 - 311 páginas |
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Página 63
... Hence , not only must we have CD < AC + AD , but also the greater radius AD AC + CD ; And whenever the triangle CAD ... Hence , if the distance between the centres of two circles be greater than the sum of their radii , the two circles ...
... Hence , not only must we have CD < AC + AD , but also the greater radius AD AC + CD ; And whenever the triangle CAD ... Hence , if the distance between the centres of two circles be greater than the sum of their radii , the two circles ...
Página 86
... hence the side PC will ap- ply to its equal PB , and the point C will fall on B ; besides , from the nature of the polygon , the angle PCD = PBA ; hence CD will take H the direction BA , and since CD = BA , the point D will fall on A ...
... hence the side PC will ap- ply to its equal PB , and the point C will fall on B ; besides , from the nature of the polygon , the angle PCD = PBA ; hence CD will take H the direction BA , and since CD = BA , the point D will fall on A ...
Página 157
... hence , if instead of the part PCDEQ , we substitute the straight line PQ , the enveloping line APQB will be shorter than APDQB . But , by hypothesis , this latter was shorter than any other ; hence that hypothesis was false ; hence all ...
... hence , if instead of the part PCDEQ , we substitute the straight line PQ , the enveloping line APQB will be shorter than APDQB . But , by hypothesis , this latter was shorter than any other ; hence that hypothesis was false ; hence all ...
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Términos y frases comunes
ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC angles equal arc AC base BC bisected centre chord circle ABC circumference cosine cylinder demonstrated diameter divided draw Prob equal and similar equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given straight line greater Hence hypotenuse inscribed join less Let ABC Let the straight magnitudes meet multiple opposite angle parallel parallelogram parallelopiped perpendicular polygon prism PROP proposition quadrilateral radius ratio rectangle contained rectilineal figure remaining angle right angled triangle SCHOLIUM segment semicircle shewn side BC sine solid angle solid parallelopiped spherical angle spherical triangle square straight line BC THEOR third touches the circle triangle ABC wherefore