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
... ABC shall be an equilateral triangle . Because the point A is the centre of the circle BCD , AC is equal ( Definition 15. ) to AB ; and because the point B is the centre of the circle ACE , BC is equal to BA : But it has been proved ...
... ABC shall be an equilateral triangle . Because the point A is the centre of the circle BCD , AC is equal ( Definition 15. ) to AB ; and because the point B is the centre of the circle ACE , BC is equal to BA : But it has been proved ...
Side 8
... triangle ABC to the A.A CE T triangle DEF ; and the other angles , to which the equal sides are opposite , shall be equal each to each , viz . the angle ABC to the angle DEF , 8 EUCLID'S ELEMENTS .
... triangle ABC to the A.A CE T triangle DEF ; and the other angles , to which the equal sides are opposite , shall be equal each to each , viz . the angle ABC to the angle DEF , 8 EUCLID'S ELEMENTS .
Side 9
... triangle ABC be applied to DEF , SO that the point A may be on D , and the straight line AB upon DE ; the point в ... triangle ABC shall coincide with the whole triangle DEF , and be equal to it ; and the other angles of the one ...
... triangle ABC be applied to DEF , SO that the point A may be on D , and the straight line AB upon DE ; the point в ... triangle ABC shall coincide with the whole triangle DEF , and be equal to it ; and the other angles of the one ...
Side 10
Euclides Thomas Tate. Let ABC be an isosceles triangle , of which the side A B is equal to a C , and let the straight lines A B , AC be produced to D and E , the angle ABC shall be equal to the angle ACB , and the angle CBD to the ...
Euclides Thomas Tate. Let ABC be an isosceles triangle , of which the side A B is equal to a C , and let the straight lines A B , AC be produced to D and E , the angle ABC shall be equal to the angle ACB , and the angle CBD to the ...
Side 11
... ABC : And it has also been proved that the angle FBC is equal to the angle GCB , which are the angles upon the other side of the base . Therefore the angles at the base , & c . Q. E. D. COROLLARY . Hence every equilateral triangle ...
... ABC : And it has also been proved that the angle FBC is equal to the angle GCB , which are the angles upon the other side of the base . Therefore the angles at the base , & c . Q. E. D. COROLLARY . Hence every equilateral triangle ...
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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