| John Keill - Logarithms - 1723 - 364 pages
...Angle HAC is alfo equal to the Angle MDF. Therefore the two Triangles MDF, HAC, have two Angles of the **one equal to two Angles Of the other, each to each, and** one Side of the one equal to one Side of the other, viz. that which is fubtended by one of the equal... | |
| John Keill - Trigonometry - 1733 - 397 pages
...Angle HAG is alfo equa to the Angle MDF. Therefore the two Triangles MDF, HA C, have two Angles of the **one equal to two Angles of the other, each to each, and** one Side of the one equal to one Side of the other, viz. that which is fubtended by one of the equal... | |
| Robert Simson, Euclid - Trigonometry - 1762 - 466 pages
...HCF is equal to KCF, and the right angle FHC equal to the right angle FKC ; in the triangles FHC, FKC **there are two angles of one equal to two angles of the other** ; and the fide FC, which is oppofite to one of the equal angles in each, is common to both ; therefore... | |
| Euclides, David Gregory - Mathematics, Greek - 1765 - 464 pages
...Clavius has alfo tranflated them into Latin. PROP. XXVI. THEO R. If two triangles have two angles of the **one equal to two angles of the other, each to each, and** one fide of the one equal to one fide of the other, either that fide which is hetween the equal angles,... | |
| Robert Simson, Euclid - Trigonometry - 1775 - 520 pages
...HCF is equal to KCF, and the right angle FHC equal to the right angle FKC; in the triangles FHC, FKC **there are two angles of one equal to two angles of the other,** and the fide FC, which is oppofite to one of the equal angles in each, is common to both ; therefore... | |
| Euclid - Geometry - 1776 - 264 pages
...EDF, it muft be greater. Wherefore, &c. PROP. XXVI. THEO R. TF two triangles have two angles of the **one equal to two angles •*• of the other, each to each, and** aJiJe of the one equal to ajide of the other, either thejide lying between the equal angks, orj'ubtending... | |
| Robert Simson - Trigonometry - 1781 - 466 pages
...CFD double of CFL i therefore the angle KFC is equal to th< -rr angle CFL ; and the right angi« \ **FCK is equal to the right angle FCL. therefore in...to two angles of the other, each to each, and the** fide FC, which is adjacent to the equal angles in each, is common to both ; therefore the other fides... | |
| Euclid - 1781 - 520 pages
...Fl angle FCK is equal to the right angle FCL : Therefore, in the two triangles FKC, FLC, there ate **two angles of one equal to two angles of the other, each** t» each, and the fide FC, which is adjacent to the equal angles in each, is common to both ; theiefore... | |
| Benjamin Donne - 1796
...remaining angle of me nwji be equal to the remaining angle of the other. THEOREM 15. If two triangles have **two angles of one equal to two angles of the other, each to each, and** one s1de of one equal to one D side side of the other, the triangles are equal in every refpcEl. —... | |
| Alexander Ingram, Robert Simson - Trigonometry - 1799 - 351 pages
...Cv.ED 84. i. b 34. i. PROP. BooK I. 54.i, PROP. XXVI. THEOR. TF two triangles have two angles of the **one equal to -*- two angles of the other, each to each ; and** one fide equal to one fide, viz. either the fides adjacent to the equal angles, or the fides oppofite... | |
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