 | Robert Simson - 1762 - 488 sider
...bafe EF. Therefore in equal circles, &c. Q._E. D. PROP. XXX. PRO B. 3 bifect a given circumference, that is to divide it into two equal parts. Let ADB be the given circumference ; it is required to bifect it. Join AB, and bifect a it in C ; from the point C draw... | |
 | John Playfair, Euclid - 1804 - 468 sider
...the bafe EF. Therefore, c 4 ». in equal circles, &c. Q^ED PROP. XXX. ^PROB. T O bifeft a given arch, that is, to divide it into two equal parts. Let ADB be the given arch ; it is required to bifedb it. Join AB,'and bifeft» it in C ; from the point C draw CD a i0.... | |
 | Robert Simson - 1804 - 530 sider
...the bafe EF. Therefore, in equal circles, &c. Q^ED PROP. XXX. PROB. TO bifeft a given circumference, that is, to divide it into two equal parts. Let ADB be the given circumference ; it is required to bifecl it. a. 10. i. J0in AB, and bifect » it in C ; from the point... | |
 | Euclides - 1816 - 588 sider
...the base EF. Therefore, in equal circles, &c. QED PROP. XXX. PROB. To bisect a given circumference, that is, to divide it into two equal parts. Let ADB be the given circumference; it is required to bisect it. ^ • 10. 1. Join AB, and bisect1 it in C ; from the point... | |
 | John Playfair - 1819 - 350 sider
...(4. 1.) to the base EF. Therefore, in equal circles, &c. QED PROP. XXX. PROB. To bisect a given arch, that is, to divide it into two equal parts. Let ADB be the given arch ; it is required to bisect it. Join AB, and bisect (10. 1.) it in C ; from the point C draw CD... | |
 | Robert Simson - 1827 - 546 sider
...EF. Therefore in *4. i• equal circles, &c. QED PROP. XXX. PROB. To bisect a given circumference, that is, to divide it into two equal parts. Let ADB be the given circumference ; it is required to bisect it. Join AB, and bisect * it in C ; from the point C draw... | |
 | John Playfair - 1835 - 336 sider
...there will be formed a right angled triangle equal to ABE (Prop. 18. BI). PROP. III. PROB. To bisect a given arc, that is, to divide it into two equal...parts. Let ADB be the given arc ; it is required to bisect it. Join AB, and bisect (Prob. 5. 1.) it in C ; from the point C draw CD at right angles to... | |
 | John Playfair - 1836 - 148 sider
...the base EF. Therefore, in equal circles, &c. QED PROP. XXL PROB. To bisect a given circumference, that is, to divide it into two equal parts. Let ADB be the given circumference ; it is required to bisect it. Join AB, and bisect (8. 1.) it in C; from the point C... | |
 | John Playfair - 1837 - 332 sider
...equal, their radii are equal : therefore BK, KG are equal to EL, LF : and 80 PROP. XXX. THEOR. To bisect a given arc, that is, to divide it into two equal...parts. Let ADB be the given arc ; it is required to bisect it. Join AB, and bisect (10. 1.) it in C ; from the point C draw CD at right angles to AB, and... | |
 | Andrew Bell - 1837 - 290 sider
...the centres of the circles (III. 1), and join BK, «: *> sx PROPOSITION XXX. PROBLEM. * ' To bisect a given arc, that is, to divide it into two equal...parts. Let ADB be the given arc ; it is required to bisect it. Join AB, and bisect it in C (1. 10) ; from the point C draw CD at right angles to AB, and... | |
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To bisect a given arc, that is, to divide it into two equal parts. Let ADB be the given arc : it is required to bisect it. 
