The first three books of Euclid's Elements of geometry, with theorems and problems, by T. Tate1851 - 139 sider |
Inni boken
Resultat 1-5 av 34
Side 10
... take any point F , and from A E , the greater , cut off AG equal ( 1. 3. ) to AF , the less , and join FC , GB . D A B C G Because AF is equal to AG , and AB to AC ( Hyp . ) , the two sides F A , AC are equal to the two GA , AB , each ...
... take any point F , and from A E , the greater , cut off AG equal ( 1. 3. ) to AF , the less , and join FC , GB . D A B C G Because AF is equal to AG , and AB to AC ( Hyp . ) , the two sides F A , AC are equal to the two GA , AB , each ...
Side 14
... Take any point D in A B , and from AC cut ( 1. 3. ) off A E equal to A D ; join D E , and upon it , on the side towards B and C , describe ( 1. 1. ) an equilateral tri- angle DEF ; then join AF ; the straight line AF bisects the ...
... Take any point D in A B , and from AC cut ( 1. 3. ) off A E equal to A D ; join D E , and upon it , on the side towards B and C , describe ( 1. 1. ) an equilateral tri- angle DEF ; then join AF ; the straight line AF bisects the ...
Side 15
... Take any point D in AD , and ( 1. 3. ) make CE equal to CD , and upon DE des- F cribe ( 1. 1. ) the equilateral tri- angle DFE , and join FC ; the straight line FC drawn from the given point c is at right angles to the given straight ...
... Take any point D in AD , and ( 1. 3. ) make CE equal to CD , and upon DE des- F cribe ( 1. 1. ) the equilateral tri- angle DFE , and join FC ; the straight line FC drawn from the given point c is at right angles to the given straight ...
Side 17
... Take any point D upon the other side of AB , from the cen- tre C , at the distance CD , de- scribe ( Post . 3. ) the circle EGF meeting AB in F , G ; and bisect ( 1. 10. ) FG in H , and join CF , CH , CG ; the straight line cH , drawn ...
... Take any point D upon the other side of AB , from the cen- tre C , at the distance CD , de- scribe ( Post . 3. ) the circle EGF meeting AB in F , G ; and bisect ( 1. 10. ) FG in H , and join CF , CH , CG ; the straight line cH , drawn ...
Side 19
... Take away the common angle ABC , the remaining angle ABE is equal ( Ax . 3. ) to the remaining angle A B D , the less to the greater , which is impossible ; therefore BE is not in the same straight line with B C. And , in like manner ...
... Take away the common angle ABC , the remaining angle ABE is equal ( Ax . 3. ) to the remaining angle A B D , the less to the greater , which is impossible ; therefore BE is not in the same straight line with B C. And , in like manner ...
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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