 ... sum of the angles at the base ABC, ACB, to the tangent of half their difference. About A as a centre, with AB the greater side for a distance, let a circle be described, meeting AC, produced in E, F, and BC in D ; join DA, EB, FB ; and draw FG parallel...  Operative Masonry: Or, A Theoretical and Practical Treatise of Building ... - Page 62by Edward Shaw - 1832 - 140 pagesFull view - About this book
 | Robert Simson - Trigonometry - 1806 - 546 pages
...and BC in D ; join DA, EB, FB ; and draw FG parallel to BC, meeting EB in G. The angle EAB (32. 1.) is equal to the sum of the angles at the base, and the angle EFB at the circumference is equal to the half of EAB at the centre (50. 3.); therefore EFB... | |
 | Euclides - 1816 - 588 pages
...F, and BC in D; join DA, EB, FB; and draw FG parallel to BC, meeting EB in G. The angle EAB (32. 1.) is equal to the sum of the angles at the base, and the angle EFB at the circumference is equal to the half of EAB at the centre (20. 3.); therefore EFB... | |
 | Sir John Leslie - Curves - 1821 - 518 pages
...equal to BAC, and produce it till BC' be equal to BC ; then AD'C, being equal to D'BC and BCD' or BAC, is equal to the sum of the angles at the base, and therefore given. Now the triangles AD'C and BD'C' are evidently similar, and consequently AD' : D'C : : DC' :... | |
 | Robert Simson - Trigonometry - 1827 - 546 pages
...and BC in D ; join DA, EB, FB ; and draw FG parallel to BC, meeting EB in G. The angle EAB (32. 1.) is equal to the sum of the angles at the base, and the angle EFB at the circumference is equal to the half of EAB at the centre (20. 3.); therefore EFB... | |
 | Peter Nicholson - Masonry - 1828 - 276 pages
...FAC, CAB, BAE, are also equal to two right angles, and since FAC is equal to the vertical angle of the triangle, the angle CAE is equal to the sum of the...point of convergency may be in the centre of the arc. Fig. 12. Let AB and EF be the two converging lines, and A the given point through which the arc is... | |
 | Euclid - 1835 - 540 pages
...and BC in D ; join DA, EB, FB : and draw FG parallel to BC, meeting EB in G. The angle EAB (32. 1.) is equal to the sum of the angles at the base, and the angle EFB at the circumference is equal to the half of EAB at the centre, (20. 3.) ; therefore... | |
 | Edward Shaw - Masonry - 1846 - 342 pages
...two right angles, and since F А С is equal to the vertical angle of the triangle, the angle С AE is equal to the sum of the angles at the base ; and therefore, since С А B is half the sum, the angle ABD must be equal to the other half. PROBLEM XII. Given two converging... | |
 | Horatio Nelson Robinson - Mathematics - 1851 - 96 pages
...difference of the angles at the base. The angle BCD, being the exterior angle of the triangle ACB, is equal to the sum of the angles at the base; and as DEB is at the circumference, and stands on the same arc, DEB is the half sum of the angles at the... | |
| |
