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 53
... Let ABC be any triangle , and the angle at B one of its acute angles , and upon BC , one of the sides containing it , let fall the perpendicular ( Prob . 7. 1. ) AD from the opposite angle : The square of AC , opposite to the an- gle B ...
... Let ABC be any triangle , and the angle at B one of its acute angles , and upon BC , one of the sides containing it , let fall the perpendicular ( Prob . 7. 1. ) AD from the opposite angle : The square of AC , opposite to the an- gle B ...
Página 146
... Let ABC be a circle , of which AC is a diameter , let DE be perpendicu- lar to the diameter AC , and let AB meet DE in F ; the rectangle BA.AF is equal to the rectangle CA.AD. Join BC , and because ABC is an an- E B F D B คน gle in a ...
... Let ABC be a circle , of which AC is a diameter , let DE be perpendicu- lar to the diameter AC , and let AB meet DE in F ; the rectangle BA.AF is equal to the rectangle CA.AD. Join BC , and because ABC is an an- E B F D B คน gle in a ...
Página 246
... Let the triangle ABCe right angled at A , and let AC be either of the sides ; the sine of the botenuse BC will be to the radius as the sine of the arc AC is to the sinof the angle ABC . Let D be the centre athe sphere , and let CE be ...
... Let the triangle ABCe right angled at A , and let AC be either of the sides ; the sine of the botenuse BC will be to the radius as the sine of the arc AC is to the sinof the angle ABC . Let D be the centre athe sphere , and let CE be ...
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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