 | Robert Gibson - 1795 - 384 sider
...lame Parallels the Angle ACE=A (by Part 2. of the laft) There.. fore ECD+ACE=ACD=B+A. Q^ED THE 0. V. In any Triangle ABC, all the three Angles taken together are equal to tiw right Angles, viz. A.\.B.\.ACB =2 right Angles. Fig. 23, Produce BC to any Diftance as D, then... | |
 | William Duncan - 1802 - 256 sider
...deduced, almost as soon *s proposed. Thus Euclid having demonstrated, that in every right-lined triangle, all the three angles taken together are equal to two right angles ; adds, by way of corollary, that all the three 'angles of any one triangle taken together, are equal... | |
 | William Duncan - 1802 - 258 sider
...deduced, almost as soon as proposed. Thus Euclid having demonstrated, that in every right-lined triangle, all the three angles taken together are equal to two right angles ; adds, by way of corollary, that all the three angles of any one triangle taken together, are equal... | |
 | Robert Gibson - 1806 - 486 sider
...parallels, the angle ACE = A (by part 2. of the last.) Therefore ECD + ACE = ACD =1 B + AQED THEOREM V. In any triangle ABC, all the three angles taken together are equal to two right angles, viz. A + B+ ACB = 2 right angles. Fig. 23. Produce CB to any distance, as D, then (by the last) ACD=B+A... | |
 | Robert Gibson - 1808 - 482 sider
...parallels, the angle ACE = A (by part 2. of the last.) Therefpre £CD + ACE = ACD = B + AQED THEO. V. f* any triangle ABC, all the three angles taken together are equal to two right angles, -viz. jl + £+ 4CB = 2 right angles, jig. 23. Produce BC to any distance, as D, then (by the last)... | |
 | Encyclopaedia Britannica - 1810 - 816 sider
...deduced, almolt as foon as propofed. Thus Euclid having demonstrated, " that in every right-lined triangle all the three angles taken together are equal to two right angles ;" adds by way of corollary, " that all the three angles of any one triangle taken together are equal... | |
 | Robert Gibson - 1811 - 580 sider
...the angle ACE = A (by part 2. of the last.) Therefore ECD+ACE = ACD=B+A. 2. ED THEO. V. PL. I./?. 23. In any triangle ABC, all the three angles, taken together, are equal to tiyo right angles, r/t. A+B+ ACB = 2 right angles. Produce CB to any distance, as D, then (bv the last)... | |
 | Robert Gibson - 1814 - 558 sider
...angle ACE = A (by part, 2. of the last.) Therefore ECD + ACE = ACD = B + A. QED THEO. V. PL. l.^sr.23. In any triangle ABC, all the three angles, taken together, are equal to two right angles, -viz. A -f. B + ACB = 2 rig/it angles. Produce CB to any distance, as D, then (by the last) ACD=B+A;... | |
 | 1816 - 746 sider
...deduced, almoft as foon as propofed. Thus Euclid having demonftrated, " that in every right-lined triangle all the three angles taken together are equal to two right angles," adds by way. of corrollary, " that all the three angles of any one triangle taken together are equal... | |
 | Robert Gibson - 1818 - 502 sider
...same parallels, ^he angle ACE=rA (by part 2. of the last.) Therefore ECD+ACE=ACD =B+AQED THEOREM V. In any triangle ABC, all the three angles taken together are equal to two right angles, viz, Jl+B+JlCB =2 right angles. fig. 23. Produce BC to any distance, as D, then (by the last.) ACD^B+A;... | |
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Thro' C, let CE be drawn parallel to AB ; then since BD cuts the two parallel lines BA, CE ; the angle ECD = B, (by part 3, of the last theo.) and again, since AC cuts the same parallels, the angle ACE = A (by part 2. of the last.) Therefore ECD + ACE... 
