Elements of Geometry and Trigonometry |
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Page 47
... . XV . ) ; hence the D point E is without the circle ; therefore , BD has no point but A common to it and the circumference ; consequently BD is a tangent ( Def . 8. ) . Scholium . At a given point A , only one BOOK III . 47.
... . XV . ) ; hence the D point E is without the circle ; therefore , BD has no point but A common to it and the circumference ; consequently BD is a tangent ( Def . 8. ) . Scholium . At a given point A , only one BOOK III . 47.
Page 69
... consequently equal ( Book I. Prop . X. ) . But if from the quadrilateral ABED , we take away the tri- angle ADF , there will remain the parallelogram ABEF ; and if from the same quadrilateral ABED , we take away the equal triangle CBE ...
... consequently equal ( Book I. Prop . X. ) . But if from the quadrilateral ABED , we take away the tri- angle ADF , there will remain the parallelogram ABEF ; and if from the same quadrilateral ABED , we take away the equal triangle CBE ...
Page 78
... consequently the triangle HBC , and the square ÄH , which have the common base BH , have also the common altitude AB ; hence the triangle is half of the square , The triangle ABF has already been proved equal to the tri- angle HBC ...
... consequently the triangle HBC , and the square ÄH , which have the common base BH , have also the common altitude AB ; hence the triangle is half of the square , The triangle ABF has already been proved equal to the tri- angle HBC ...
Page 80
... consequently BD2 - BC2 + CD2—2BC × CD ( Prop . IX . ) . Adding AD2 to each , and observing that the right angled trian- gles ABD , ADC , give AD2 + BD2 = AB2 , and AD2 + CD2 AC2 , we have AB2 = BC2 + AC2-2BC x CD . Secondly . When the ...
... consequently BD2 - BC2 + CD2—2BC × CD ( Prop . IX . ) . Adding AD2 to each , and observing that the right angled trian- gles ABD , ADC , give AD2 + BD2 = AB2 , and AD2 + CD2 AC2 , we have AB2 = BC2 + AC2-2BC x CD . Secondly . When the ...
Page 81
... consequently the triangle ABC gives AB2 + BC - 2AE2 + 2BE2 . The triangle ADC gives , in like manner . AD2 + DC - 2AE2 + 2DE2 . E Adding the corresponding members together , and observing that BE and DE are equal , we shall have AB2 + ...
... consequently the triangle ABC gives AB2 + BC - 2AE2 + 2BE2 . The triangle ADC gives , in like manner . AD2 + DC - 2AE2 + 2DE2 . E Adding the corresponding members together , and observing that BE and DE are equal , we shall have AB2 + ...
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Common terms and phrases
adjacent adjacent angles altitude angle ACB angle BAC ar.-comp base multiplied bisect Book centre chord circ circumference circumscribed common cone consequently convex surface cylinder diagonal diameter dicular distance draw drawn equal angles equally distant equation equiangular equivalent figure formed four right angles frustum given angle given line gles greater homologous sides hypothenuse inscribed circle inscribed polygon intersection less Let ABC let fall logarithm measured by half number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedron polygon ABCDE prism proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle S-ABC Scholium secant secant line segment side BC similar solid angle solid described sphere spherical polygon spherical triangle square described straight line tang tangent THEOREM triangle ABC triangular prism vertex