The Synoptical Euclid; Being the First Four Books of Euclid's Elements of Geometry from the Edition of Dr. Robert Simson ... With Exercises. By S. A. Good ... Second Edition |
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Αποτελέσματα 1 - 5 από τα 42.
Σελίδα 7
Let AB be the given straight line ; it is required to describe an equilateral triangle upon it . From the centre A , at the ... AC is equal to AB ; and because the point B is the centre of the circle ACE , 2 . BC is equal to BA .
Let AB be the given straight line ; it is required to describe an equilateral triangle upon it . From the centre A , at the ... AC is equal to AB ; and because the point B is the centre of the circle ACE , 2 . BC is equal to BA .
Σελίδα 8
The straight line AL is equal to BC . Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC . Which was to be done . PROP . III . - PROBLEM , From the greater of two given straight lines ...
The straight line AL is equal to BC . Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC . Which was to be done . PROP . III . - PROBLEM , From the greater of two given straight lines ...
Σελίδα 9
Let ABC , DEF be two triangles which have the two sides AB , AC equal to the two sides DE , DF , cach to each , viz . AB to DE , and AC to ... The point C shall coincide with the point F , because the straight line AC is equal to DF .
Let ABC , DEF be two triangles which have the two sides AB , AC equal to the two sides DE , DF , cach to each , viz . AB to DE , and AC to ... The point C shall coincide with the point F , because the straight line AC is equal to DF .
Σελίδα 13
AB to DE , and AC to DF ; and also the base BC equal to the base EF . The angle BAC is equal to the angle EDF . D G А F B For if the triangle ABC be applied to DEF , so that the point B be on E , and the straight line BC upon EF ; 1 .
AB to DE , and AC to DF ; and also the base BC equal to the base EF . The angle BAC is equal to the angle EDF . D G А F B For if the triangle ABC be applied to DEF , so that the point B be on E , and the straight line BC upon EF ; 1 .
Σελίδα 14
upon it an equilateral triangle ABC , and bisect ( I. 9. ) the angle ACB by the straight line CD . AB is cut into two equal parts in the point D. ) C A. B Because AC is equal to CB , and CD common to the two triangles ACD , BCD ; 1 .
upon it an equilateral triangle ABC , and bisect ( I. 9. ) the angle ACB by the straight line CD . AB is cut into two equal parts in the point D. ) C A. B Because AC is equal to CB , and CD common to the two triangles ACD , BCD ; 1 .
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AB is equal ABC is equal ABCD angle ABC angle ACB angle AGH angle BAC angle BCD angle equal base BC bisected centre circle ABC circumference coincide common demonstrated describe diameter distance divided double draw equal angles exterior angle extremity fall figure four given circle given point given straight line gnomon greater impossible inscribed join less Let ABC Let the straight likewise manner meet opposite angles parallel parallelogram pass pentagon perpendicular point F PROBLEM produced Q.E.D. PROP reason rectangle contained rectilineal figure remaining angle right angles segment semicircle shown side BC sides square of AC straight line AC Take touches the circle triangle ABC twice the rectangle wherefore whole