The first three books of Euclid's Elements of geometry, with theorems and problems, by T. Tate1851 - 139 sider |
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Side 6
Euclides Thomas Tate. XII . " If a straight line meets two straight lines ... AC is equal ( Definition 15. ) to AB ; and because the point B is the centre ... straight line AB . Which was required to be done . CB ; PROP . II . PROB . From a ...
Euclides Thomas Tate. XII . " If a straight line meets two straight lines ... AC is equal ( Definition 15. ) to AB ; and because the point B is the centre ... straight line AB . Which was required to be done . CB ; PROP . II . PROB . From a ...
Side 8
... straight line AD equal to C ; and E B from the centre A , and at the distance AD , describe ( Post . 3. ) the circle ... AC equal to the two sides DE , DF , each to each , A viz . AB to DE , and AC to DF , and the angle BAC equal to the ...
... straight line AD equal to C ; and E B from the centre A , and at the distance AD , describe ( Post . 3. ) the circle ... AC equal to the two sides DE , DF , each to each , A viz . AB to DE , and AC to DF , and the angle BAC equal to the ...
Side 9
... A may be on D , and the straight line AB upon DE ; the point в shall coincide with the point E , because AB is equal to DE ( Hyp . ) ; and AB coinciding with DE , AC shall coincide with DF because the angle BAC is equal to the angle EDF ...
... A may be on D , and the straight line AB upon DE ; the point в shall coincide with the point E , because AB is equal to DE ( Hyp . ) ; and AB coinciding with DE , AC shall coincide with DF because the angle BAC is equal to the angle EDF ...
Side 13
... A C , equal to the two sides , DE , DF , each to each , viz . AB to DE , and A AC to DF , and also the base BC equal ... straight line BC upon EF ; the point C shall also coincide with the point BC because BC is equal to EF ( Hyp . ) ; ...
... A C , equal to the two sides , DE , DF , each to each , viz . AB to DE , and A AC to DF , and also the base BC equal ... straight line BC upon EF ; the point C shall also coincide with the point BC because BC is equal to EF ( Hyp . ) ; ...
Side 14
... AC cannot but coincide with the sides E D , D F , wherefore likewise the ... straight line AF bisects the triangle BA C. Because A D is equal to A E ... straight line a F. Which was to be done . PROP . X. PROB . To bisect a given finite ...
... AC cannot but coincide with the sides E D , D F , wherefore likewise the ... straight line AF bisects the triangle BA C. Because A D is equal to A E ... straight line a F. Which was to be done . PROP . X. PROB . To bisect a given finite ...
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Vanlige uttrykk og setninger
A B C ABCD adjacent angles angle ABC angle ACB angle BAC angle equal angles BGH base BC BC is equal bisect centre circle ABC circumference diameter divided double draw a straight equal angles equal circles equal straight lines equal to FB exterior angle given point given rectilineal angle given straight line gnomon greater half a right hypotenuse interior and opposite isosceles triangle less Let ABC Let the straight line be drawn opposite angles parallel parallelogram perpendicular PROB produced Q. E. D. PROP rectangle AE rectangle contained rectilineal figure remaining angle right angles segment semicircle side BC square of AC straight line AB straight line AC straight line drawn THEOR touches the circle trapezium triangle ABC twice the rectangle vertex vertical angle