| John Keill - Logarithms - 1723 - 364 pages
...every of the Angles KHG,HGM,GML, is equal to the Angle HKL, or KLM. Therefore the fiveAngles GHK,HKL, **KLM, LMG, MGH, are equal between themfelves. And fo...equilateral, and defcribed about the Circle ABCDE** j •which was to be dune. PROPOSITION XIII, PROBLEM. To tinfcribe a Circle in an equilateral and equiangular... | |
| Robert Simson - Trigonometry - 1762 - 466 pages
...before demonftrated ; the angle HKL is equal to KLM. and in like manner it may be fliewn, that each **of the angles KHG, HGM, GML is equal to the angle...therefore the five angles GHK, HKL, KLM, LMG, MGH** being equal to one another, the pentagon GHKLM is equiangular. and it is equilateral, as was demonftrated... | |
| Euclid, Edmund Stone - Geometry - 1765 - 464 pages
...We demonftrate after the fame manner that the angles KHG, HGM, GML are each equal to the angles HKL, **KLM: Therefore the five angles GHK, HKL, KLM, LMG, MGH are equal** to one another. Wherefore the I pentagon GHK t M is equiangular. But it has alfo been proved to be... | |
| Robert Simson - Trigonometry - 1775 - 520 pages
...the angle HKL is equal to KLM : And in like manner it may be fhewn, that each of the angles KHG, HGMi **GML is equal to the angle HKL or KLM : Therefore the five angles GHK, HKL, KLM, LMG, MGH** being equal to one another, the pentagon GHKLM is equiangular : And it is et;uil.ttcr.tl, as was demonftrated... | |
| Euclid, James Williamson - Euclid's Elements - 1781 - 309 pages
...manner each of the angles KHG, HGM, GML will be demonftrated to be equal to either of the angles HKL, **KLM; therefore the five angles GHK, HKL, KLM, LMG, MGH are equal** to one another : therefore the pentagon GHKLM is equiangular ; but it has been alfo demonftrated to... | |
| Euclid, John Playfair - Electronic book - 1795 - 400 pages
...before demonftrated, the angle HKL is equal to KLM : and in like manner it may be fliown, that each **of the angles KHG, HGM, GML is equal to the angle...therefore the five angles GHK, HKL, KLM, LMG, MGH** being equal to one another, the pentagon GHKLM is equiangular : and it is equilateral, as was demouftrated... | |
| Robert Simson - Trigonometry - 1804
...was before demonftrated; the angle HKL is equal to KLM. and in like manner it may be fhewn, that each **of the angles KHG, HGM, GML is equal to the angle...therefore the five angles GHK, HKL, KLM, LMG, MGH** being equal to one another, the pentagon GHKLM is equiangular, and it is equilateral, as was demonftrated;... | |
| Robert Simson - Trigonometry - 1806 - 518 pages
...angle HKL is equal to KLM: and in like manner it may be shown, that each of the angles KHG, ! IG.M, **GML is equal to the angle HKL or KLM : therefore the five angles GHK, HKL, KLM, LMG, MGH** being equal to one another, the pentagon GHKLM is equiangular : and it is equilateral, as -was demonstrated... | |
| John Playfair - Euclid's Elements - 1806 - 311 pages
...as was before demonstrated, the angle HKL is equal to KLM. In like manner it may be shown that each **of the angles KHG, HGM, GML is equal to the angle HKL,** of KLM. Therefore the Book IV. five angles GHK, HKL, KLM, LMG, MGH being equal to one another, the... | |
| Euclides - 1816 - 528 pages
...before demonstrated, the angle HKL is equal to KLM : And in like manner it may be shown, that each **of the angles KHG, HGM, GML is equal to the angle...Therefore the five angles GHK, HKL, KLM, LMG, MGH,** being tqual to one another, the pentagon GHKLM is equiangular: And it is equilateral, as was demonstrated;... | |
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