| Robert Simson, Euclid - Trigonometry - 1762 - 466 pages
...ferences. • Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centers, and BAC, **EDF at their circumferences, the circumference BKC...centers are equal ; therefore the two fides BG; GC,** arc equal to the two EH, HF; and the angle at G is equal to the angle at H; therefore the bafe BC is... | |
| Euclides, David Gregory - Mathematics, Greek - 1765 - 464 pages
...drawn from their centres will be equal : Therefore the two fides BG, G c are equal to the two fides **EH, HF, and the angle at G is equal to the angle at H : Therefore** [by 4. i.] the bi,fe B c is equal to the bafe B F. And becaufe the angle at A is equal to the angle... | |
| Robert Simson, Euclid - Trigonometry - 1775 - 520 pages
...circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centers, and BAC, **EDF at their circumferences : The circumference BKC...drawn from their centers are equal : therefore the two** ftdes BG, GC, are equal to the two EH, HF ; and the angle at G is equal to the angle at II ; therefore... | |
| Robert Simson - Trigonometry - 1781 - 466 pages
...circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHFat their centers, and BAC, **EDF at their circumferences. the circumference BKC...fides BG, GC, are equal to the two EH, HF ; and the** angte 'CE at G is equal to the angle at H ; therefore the bafe BC is equal ' >, 4- <• to the bafe... | |
| Euclid - 1781 - 520 pages
...circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, **EDF at their circumferences : The circumference BKC...DEF are equal, the ftraight lines drawn from their** centres are equal : Theretore the two fides BG, GC, are equal to the two EH, I IF ; and the angle at... | |
| John Playfair - Euclid's Elements - 1795 - 400 pages
...circumferences. .-'let Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, **EDF at their circumferences : the circumference BKC...Join BC, EF< and becaufe the circles ABC, DEF are** eqtial, the ftraXght Ijnes dijwvn from, their centres are «qual: therefore the two fides, ^G, GC,... | |
| Alexander Ingram, Robert Simson - Trigonometry - 1799 - 351 pages
...and becaufe the circles ABC, DEF are equal, *— '"V^^ the flraight lines drawn from their centres **are equal : Therefore the two fides BG, GC are equal...angle at G is equal to the angle at H ; therefore the** a 4. i. bafe BC is equal a to the bafe EF : And becaufe the angle at A lIn.def-3. 's equal to the angle... | |
| John Playfair - Euclid's Elements - 1804 - 440 pages
...Join BC, EF ; and becaufe th$ circles ABC, DEF are equal, the ftraight lines drawn from their centres **are equal: therefore the two fides BG, GC, are equal...HF; and the angle at G is equal to the angle at H** j there • 4.i. fore the bafe BC is equal » to the bafe EF : and becaufe the angle at A is equal... | |
| John Playfair - Trigonometry - 1806 - 311 pages
...ABC, DEF are equal, the straight lines drawn from their centres are equal j therefore the two sides **BG, GC are equal to the two EH, HF ; and the angle at G is equal to the angle at H** ; therea 4. 1. fore the base BC is equal* to the base EF. Because the angle at A is equal to the angle... | |
| Euclid, Robert Simson - Euclid's Elements - 1806 - 518 pages
...circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, **EDF at their circumferences : the circumference BKC...is equal to the circumference ELF. Join BC, EF; and** because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal : therefore... | |
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