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 96
... manner , it may be shewn that GH , GM , ML are each of them equal to HK or KL : therefore the pentagon GHKLM is equilateral . It is also equiangular ; for , since the angle FKC is equal to the angle FLC , and the angle HKL double of the ...
... manner , it may be shewn that GH , GM , ML are each of them equal to HK or KL : therefore the pentagon GHKLM is equilateral . It is also equiangular ; for , since the angle FKC is equal to the angle FLC , and the angle HKL double of the ...
Página 97
... manner as in the preceding proposition , that the angles CBA , BAE , B AED are bisected by the straight lines FB , FA , FE and because that the angle BCD is equal to the angle CDE , and that FCD is the half of the angle BCD , and CDF ...
... manner as in the preceding proposition , that the angles CBA , BAE , B AED are bisected by the straight lines FB , FA , FE and because that the angle BCD is equal to the angle CDE , and that FCD is the half of the angle BCD , and CDF ...
Página 109
... manner , if mA = nB , mA + mC + mE = nB ÷ nD + nF ; and if mA nB , mA + mC + mE / nB ÷ nD + nF . Now , mA + mC + mE = m ( A + C + E ) ( Cor . 1. 5. ) , so that mA and mA + mC + mE are any equimultiples of A , and of A + C + E . And for ...
... manner , if mA = nB , mA + mC + mE = nB ÷ nD + nF ; and if mA nB , mA + mC + mE / nB ÷ nD + nF . Now , mA + mC + mE = m ( A + C + E ) ( Cor . 1. 5. ) , so that mA and mA + mC + mE are any equimultiples of A , and of A + C + E . And for ...
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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