 | John Keill - Logarithms - 1723 - 444 pages
...is * equal to the whole Angle FBC. And *^.. ^ fince the two Sides AB, BD, are equal to the two Sides FB, BC, each to each, and the Angle DBA equal to the Angle FBC; the Bafe AD will be ft equal to the Bafe FC, and the Triangle ABD equal to the Triangle FBC: But the... | |
 | Euclid, John Keill - Geometry - 1733 - 444 pages
...is * equal to the whole Angle FB C. And* Ax. fince the two Sides AB, BD, are equal to the two Sides FB, BC, each to each, and the Angle DBA equal to the Angle FBC; the Bafe AD will be ft 4 equaj to the Bafe FC, and the Triangle ABD equal to the Triangle FBC: But... | |
 | Robert Simson - Trigonometry - 1762 - 488 pages
...the fame reafon CF is equal to FD. and becaufe AD is equal to BC, and AF to FB, the two fides FA, AD are equal to the two FB, BC, each to each ; and the bafe DF was proved equal to the bafe FC ; therefore the angle FAD is eeq'.ial d to the angle FBC. agam,... | |
 | Robert Simson - Trigonometry - 1775 - 534 pages
...ABC, and the whole angle DBA is eq'ualc to the whole FBC ; and becaufe the two fides AB, BD c •- Ax. are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC ; therefore the bafe AD is eqwlftothe bafe FC, and the triangle ABD to the triangle f 4- i. IBC : Now the parallelogram... | |
 | Euclid - 1781 - 552 pages
...fame reafon, CF is equal to FD : And becaufe AI? is equal to BC, and AF to FB, the two fides FA, AD are equal to the two FB, BC, each to each ; and the bafe DF was proved equal to the bafe FC ; therefore the angle FAD is equal d to the angle FBC : Again,... | |
 | Euclid, John Playfair - Euclid's Elements - 1795 - 462 pages
...and the whole angle DBA is equal e to til whole FBC ; and becaufe the two fides AB, BD are equal tt the two FB, BC, each to each, and the angle DBA equal th the angle FBC ; therefore the bafe AD is equal f to the bafe Book I. FC, and the triangle \BD to... | |
 | Alexander Ingram - Trigonometry - 1799 - 374 pages
...angle ARC, e s. Ax. and tne whole angle DBA is equal e to the whole FBC ; and becaufe the two fides AB, BD are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC ; f 4. i. therefore the bafe AD is equal i to the bafe FC, and the triangle ABD to the triangle FBC... | |
 | John Playfair, Euclid - Circle-squaring - 1804 - 468 pages
...each the angle ABC, the whole angle DBA will be equal e to the whole FBC ; and becaufe the two fides AB, BD, are equal to the two FB, BC, each to each, and the angle DBA equal to tho the angle FBC, therefors the bafe AD is equal f te the bafe Book I. FC, and the triangle ABD to... | |
 | Robert Simson - Trigonometry - 1804 - 530 pages
...the fame reafon CF is equal to FD. and becaufe AD is equal to BC, and AF to FB, the two fides FA, AD are equal to the two FB, BC, each to each; and the bafe DF was proved equal to the bafe FC ; therefore the angle FAD is equal d to the angle FBC. again,... | |
 | John Playfair - Mathematics - 1806 - 320 pages
...of them being a right angle, add to each the angle ABC, and the whole H D angle DBA will be equalf to the whole FBC ; and because the two sides AB, BD...DBA equal to the angle FBC, therefore the base AD is equals to the base FC, and the triangle ABD to the triangle FBC. But the parallelogram BL is double11... | |
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FBC ; and because the two sides AB, BD are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the base AD is equal (i. 
