The Synoptical Euclid; Being the First Four Books of Euclid's Elements of Geometry, from the Edition of Dr. Robert Simson ... With ExercisesCharles Henry Law, 1854 - 120 σελίδες |
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Σελίδα 3
... semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter . XIX . " A segment of a circle is the figure contained by a straight line , and the circumference it cuts off . " XX ...
... semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter . XIX . " A segment of a circle is the figure contained by a straight line , and the circumference it cuts off . " XX ...
Σελίδα 62
... semicircle BHF , and produce DE to H : the square described upon EH shall be equal to the given rectilineal figure 4 . H B G E Join GH ; and because the straight line BF is 62 EUCLID'S ELEMENTS . to each of these equals add the square ...
... semicircle BHF , and produce DE to H : the square described upon EH shall be equal to the given rectilineal figure 4 . H B G E Join GH ; and because the straight line BF is 62 EUCLID'S ELEMENTS . to each of these equals add the square ...
Σελίδα 80
... semicircle is said to be greater than any acute rectilineal angle , and the remaining angle less than any rectilineal angle . ” COR . From this it is manifest that the straight line which is drawn at right angles to the diameter of a ...
... semicircle is said to be greater than any acute rectilineal angle , and the remaining angle less than any rectilineal angle . ” COR . From this it is manifest that the straight line which is drawn at right angles to the diameter of a ...
Σελίδα 84
... semicircle . Take ( III . 1. ) F the centre of the circle ABCD , and join BF , FD . And because the angle RFD is at ... semicircle . A E B F Draw AF to the centre , and produce it to C , and join CE : therefore the segment BADC is ...
... semicircle . Take ( III . 1. ) F the centre of the circle ABCD , and join BF , FD . And because the angle RFD is at ... semicircle . A E B F Draw AF to the centre , and produce it to C , and join CE : therefore the segment BADC is ...
Σελίδα 85
... semicircle , 2. The angles CAD , CED , are equal : therefore ( Ax . 2. ) 3. The whole angle BAD is equal to the whole angle BED . Wherefore the angles in the same segment , & c . Q.E.D. PROP . XXII . - THEOREM . The opposite angles of ...
... semicircle , 2. The angles CAD , CED , are equal : therefore ( Ax . 2. ) 3. The whole angle BAD is equal to the whole angle BED . Wherefore the angles in the same segment , & c . Q.E.D. PROP . XXII . - THEOREM . The opposite angles of ...
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AB is equal AC is equal adjacent angles angle ABC angle ACB angle AGH angle BAC angle BCD angle EDF angle equal base BC bisected circle ABC circumference diameter double draw equal angles equal Constr equal Hyp equal straight lines equal to BC equilateral and equiangular EUCLID'S ELEMENTS exterior angle given circle given point given rectilineal angle given straight line given triangle gnomon greater inscribed interior and opposite less Let ABC Let the straight likewise opposite angles parallel to CD parallelogram pentagon perpendicular point F produced Q.E.D. PROP rectangle AE rectangle contained remaining angle required to describe right angles semicircle side BC square of AC straight line AB straight line AC straight line drawn touches the circle triangle ABC twice the rectangle