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 17
... angle BAC : and make DG equal to AC or DF , join EG , GF . Because AB is equal to DE , and AC to DG , the two sides BA , AC are equal to the two ED , DG , each to each , and the an- gle BAC is equal to the angle EDG , therefore the base ...
... angle BAC : and make DG equal to AC or DF , join EG , GF . Because AB is equal to DE , and AC to DG , the two sides BA , AC are equal to the two ED , DG , each to each , and the an- gle BAC is equal to the angle EDG , therefore the base ...
Página 70
... angle at the centre of a circle is double of the angle at the circumference , the same base , that is , upon the same part of the circumference . upon Let ABC be a circle , and BDC an angle at the centre , and BAC an an- gle at the ...
... angle at the centre of a circle is double of the angle at the circumference , the same base , that is , upon the same part of the circumference . upon Let ABC be a circle , and BDC an angle at the centre , and BAC an an- gle at the ...
Página 141
... angle DCE : And the angle BAC was proved to be equal to ACD : Therefore the whole angle ACE is equal to the two angles ABC , BAC ; add the common angle ACB , then the angles ACE , ACB are equal to the angles ABC , BAC , ACB : But ABC , BAC ...
... angle DCE : And the angle BAC was proved to be equal to ACD : Therefore the whole angle ACE is equal to the two angles ABC , BAC ; add the common angle ACB , then the angles ACE , ACB are equal to the angles ABC , BAC , ACB : But ABC , BAC ...
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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