 | John Playfair - Circle-squaring - 1819 - 350 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 equal to one another, the pentagon GHKLM is equiangular: and it is equilateral, ae was demonstrated... | |
 | John Playfair - 1819 - 354 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 HKL or KLM : therefore the five anglea GHK, HKL, KLM, LMG, MGH being equal to one another, the pentagon GHKLM is equiangular : and... | |
 | Peter Nicholson - Mathematics - 1825 - 1046 pages
...was before demonstrated, the angle HKL is equal to KLM : And in like manner it may be shewn 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 demonstrated... | |
 | Euclid, John Playfair - Euclid's Elements - 1826 - 326 pages
...before demonstrated, the angle HKL is equal to KLM : and in like manner it may be shown, that eaeh of the angles KHG, HGM, GM-L- 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... | |
 | Euclid - 1826 - 234 pages
...manner it may be demonstrated that each of the angles KHG, HGM, GML, is equal to each 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 shown to be equilateral,... | |
 | Euclides - 1826 - 226 pages
...manner it may be demonstrated that each of the angles KHG, HGM, GML, is equal to each 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 shown to be equilateral,... | |
 | Robert Simson - Trigonometry - 1827 - 546 pages
...demonstrated, therefore the angle HKL is equal-fto KLM : and in like manner it may be shewn, 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 demonstrated... | |
 | Euclid - 1835 - 540 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 equal to one another, the pentagon GHKLM is equiangular : And it is equilateral, as was demonstrated... | |
 | Andrew Bell - Euclid's Elements - 1837 - 290 pages
...was 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 HKL or KLM ; therefore the fire angles GHK, HKL, KLM, LMG, MGH, being equal to one another, the pentagon GHKLM is equiangular... | |
 | Euclides - 1838 - 264 pages
...demonstrated ; therefore the angle HKL is equalf 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 equal to one another, the pentagon GHKLM is equiangular: and it is equilateral, as was demonstrated... | |
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And in like manner it may be shown that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM ; therefore the five angles GHK, HKL, KLM, LMG, MGH... 
