 | Robert Simson - Trigonometry - 1781 - 534 pages
...each, and the angle DBA equal to the angle FBC; therefore the bale AD is equal f to the bafe FC, f and the triangle ABD to the triangle FBC. now the parallelogram BL is double « of the triangle ABD, becaufe they are upon the s- 4ifeme bafe BD, and between the fame parallels BD, AL; and the fquare... | |
 | Alexander Ingram - Trigonometry - 1799 - 374 pages
...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 : Now the parallelogram BL is double £4i.r. g of the triangle ABD, becaufe they are upon the fame bafe BD, and between the fame parallels,... | |
 | Robert Simson - Trigonometry - 1804 - 530 pages
...each, and the angle DBA equal to the angle FBC ;. therefore the bafe AD is equal f to the bafe FC, »nd the triangle ABD to the triangle FBC. now the parallelogram BL is double f of the triangle ABD, becaufe they are upon the fame bafe BD, and between the fame parallels BD,,... | |
 | John Playfair - Mathematics - 1806 - 320 pages
...two FB, BC, each to each, and the angle 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 of the triangle ABD, and the square GB is double of the triangle... | |
 | John Playfair - 1819 - 354 pages
...the base PC, and the triangle ABD to the triangle FBC. But the parallelogram BL is double (41. 1.) of the triangle ABD, because they are upon the same base BD, and-between the same parallels, BD, AL ; and the square GB is double of the triangle BFC, because these... | |
 | John Playfair - Circle-squaring - 1819 - 350 pages
...BC each 'to each, and the ans;le DBA eqnal to the angle FBC, therefore the base AD is equal (4. 1.) to the base FC, and the triangle ABD to the triangle FBC. But the parallelogram BL is double (41. 1.) of the triangle ABD, because they are upon the same base... | |
 | Euclid, Robert Simson - Geometry - 1821 - 514 pages
...BC, each to each, and the angle DBA equal to the angle FBC; therefore thi base AD is equal (4, 1.) to the base FC, and the triangle ABD to the triangle FBC: now the parallelogram BL is double '41. 't ) of the triangle ABD, because they are upon the same base Bl), and between the same parallels... | |
 | Peter Nicholson - Mathematics - 1825 - 1046 pages
...FB, BC each to each, and the angle DBA equal to the angle FBC; therefore the base AD is equal (4. 1.) to the base FC, and the triangle ABD to the triangle FBC : Now the parallelogram BL is double (41. 1.) of the triangle ABD, because they are upon the same base BD, and between the same paralléis,... | |
 | Euclid - 1835 - 540 pages
...two FB, BC, each to each, and the angle DBA equal to the angle FBC ; therefore the base AD is equal f to the base FC, and the triangle ABD to the triangle FBC : Now the parallelogram BL is double g the triangle ABD, because they are upon the same base BD, and between the same parallels BD, AL ;... | |
 | Andrew Bell - Euclid's Elements - 1837 - 290 pages
...are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC, therefore (I. 4) the base AD is equal to the base FC, and the triangle...the parallelogram BL is double of the triangle ABD (1. 41), because they are upon the same base BD, and between the same parallels BD, AL ; and the square... | |
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FC, and the triangle ABD to the triangle FBC. Now the parallelogram BL is double... 
