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 Elements of Plane and Spherical Trigonometry |
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Página 66
IV . And the straight line on which the greater perpendicular falls , is said to be farther from the centre B. An arch of a circle is any part of the circumference . V. A segment of a circle is the figure contained by a straight line ...
IV . And the straight line on which the greater perpendicular falls , is said to be farther from the centre B. An arch of a circle is any part of the circumference . V. A segment of a circle is the figure contained by a straight line ...
Página 67
An angle in a segment is the angle contained by two straight lines drawn from any point in the circumference of the segment , to the extremities of the straight line which is the base of the segment . VII .
An angle in a segment is the angle contained by two straight lines drawn from any point in the circumference of the segment , to the extremities of the straight line which is the base of the segment . VII .
Página 80
The angles in the same segment of a circle are equal to one another . Let ABCD be a circle , and BAD , BED angles in the game segment BAED : The angles BAD , BED are equal E to one another . A Take F the centre of the circle ABCD : And ...
The angles in the same segment of a circle are equal to one another . Let ABCD be a circle , and BAD , BED angles in the game segment BAED : The angles BAD , BED are equal E to one another . A Take F the centre of the circle ABCD : And ...
Página 81
But if the segment BAED be not greater E than a semicircle , let BAD , BED be aogles jo it ; these also are equal to one another . Draw AF to the centre , and produce B D to C , and join CE : Therefore the segment BADC is greater than a ...
But if the segment BAED be not greater E than a semicircle , let BAD , BED be aogles jo it ; these also are equal to one another . Draw AF to the centre , and produce B D to C , and join CE : Therefore the segment BADC is greater than a ...
Página 82
one of the segments must therefore fall within the other : let ACB fall within ADB , draw the straight line BCD , and join CA , DA : and because the segment ACB is similar to the segment ADB , and similar segments of circles contain ...
one of the segments must therefore fall within the other : let ACB fall within ADB , draw the straight line BCD , and join CA , DA : and because the segment ACB is similar to the segment ADB , and similar segments of circles contain ...
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ABC is equal ABCD altitude angle ABC angle BAC arch base bisected Book called centre circle circle ABC circumference coincide common cylinder definition demonstrated described diameter difference divided double draw drawn equal equal angles equiangular Euclid exterior angle extremity fall fore four fourth given given straight line greater half inscribed interior join less Let ABC magnitudes manner meet multiple opposite parallel parallelogram pass perpendicular plane polygon prism produced PROP proportionals proposition proved radius ratio reason rectangle contained rectilineal figure right angles segment shewn sides similar sine solid square straight line taken tangent THEOR thing third touches triangle ABC wherefore whole