| Royal Military Academy, Woolwich - 1853
...base EF. Therefore, in equal circles, etc. QED PROPOSITION XXX. PBOB. 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 (10. i.) it in C ; from the point... | |
| Thomas Lund - 1854 - 192 páginas
...angles opposite to the equal sides AB, CD. 60. PROP. XII. To bisect a given arc of a circle. Let ABC **be the given arc. It is required to divide it into two** parts in the point B, so that the arc AB = arc EC. Join AC; bisect AC in D; from D draw DB at right... | |
| Euclides - 1855
...la the same circle, equal arcs are subtended by equal straight lines. PROP. 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 (I. 10) it at С. From the point С, draw СD at right angles (I. 11)... | |
| John Playfair - 1855 - 318 páginas
...equal angles ; therefore the base BC is equal (4. 1.) to th« base EF. PROP. XXX. THEOR. To bisect **a given arc, that is, to divide it into two equal parts. Let** ABB be the given arc ; it is required to bisect it. Join AB, and bisect (10. 1.) it in • ; from the... | |
| Robert Potts - 1860 - 361 páginas
...equal to the base EF. (I. 4.) Therefore, in equal circles, &c. QED 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. D Join AB, and bisect it in C; (i. 10.) from the point C draw CD at right angles to AB.... | |
| Euclides - 1860
...substituted ' in the same circle,' and they are often so applied. PROPOSITION XXX. PROBLEM. To bisect **a given arc ; that is, to divide it into two equal parts.** Given the arc ADC ; it is required to bisect it. (Const.) Join AB, and bisect it in C (I. 10) ; from... | |
| Euclides - 1862
...two sides EL, LF, each to each; and they contain equal angles; PROPOSITION 30.— PROBLEM. To bisect **a given arc, that is, to divide it into two equal parts.** (Kef erences— Prop. I. 4, 10, 11; III. 1 cor., 28.) Given. — Let ADB be the given arc. Sought.... | |
| Euclides - 1864
...equal to the base EF. (l. 4.) Therefore, in equal circles, &c. QED 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. D Join AB, and bisect it in C; (l. 10.) from the point Cdraw CD at right angles to AB. (T.... | |
| Euclides - 1864
...base EF. (I. 4.) Therefore, in equal circles, &c. QED PROPOSITION XXX. PROBLEM. To bisect a given are, **that is, to divide it into two equal parts. Let ADB be the given arc : it is required to** bisect it. D Join AB, and bisect it in C; (I. 10.) from the point Cdraw CD at right angles to AB. (T.... | |
| Euclides - 1865
...without or in the circumference, which shall be a tangent to a given circle . HI. 17. 3. To bisect **a given arc, that is, to divide it into two equal parts** . HE. 30. 4. A segment of a circle being given, to describe the circle of which it is the segment ......... | |
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