 | Robert Gibson - Surveying - 1806 - 486 pages
...ABCD— BDEF ; in like manner we may prove the parallelogram EFGH = BDEF. Wherefore ABDC=BDEF=EFGH. QED Cor. Hence it is plain that triangles on the same or equal bases, artd between the same parallels, are equal, seeing (by cor 2. theo. 12.) they are the halves of their... | |
 | Robert Gibson - Surveying - 1811 - 580 pages
...BDEF ; in like manner we may prove the |ğrallelogram E FG H= B DEF. Wherefore ABDC=> BDEF=EFHG. 3. ED Cor. Hence it is plain that triangles on the same...bases, and between the same parallels, are equal, seoing (by cor. 2. theo. 1Q.) theyavf the halves of their respective parallelogram • THEO. XIV. PL.... | |
 | Robert Gibson - Surveying - 1814 - 558 pages
...ABCD=BDEF ; in like manner we may prove the parallelogram EFGH=BDEF. Wherefore ABDC=BDEF=EFHG. QED ч Cor. Hence it is plain that triangles on the same...theo. 12.) they are the halves of their respective parallelogram. THEO. XIV. PL. \,ßg. 32. In every right-angled triangle, ЛВС, the square of the... | |
 | Robert Gibson - Surveying - 1818 - 502 pages
...proved that ABCD being a parallelogram, AB will be=CD and AD=BC. THEOREM XIII. •All parallelograms on the same or equal bases and between the same parallels, are equal to one another, that is, if BD=GH, and the lines BH and AF parallel, then the parallelogram J1BDC=BDFE=EFHG.... | |
 | Robert Gibson - Surveying - 1821 - 594 pages
...being a parallelogram, AB will be = CD and AD ~ BC. THEO. XIII. PL. I. Jig. 31. All parallelograma on the same or equal bases and between the same parallels, are equal to one another, that is, if BD = GH, and the lines £H and AF parallel, then the paralleloe-ram JIBDC=,BDFE.=... | |
 | Pierce Morton - Geometry - 1830 - 584 pages
...its base and altitude . cor. 16 •и two Greek words, signifying " along one (i) Parallelograms upon the same or equal bases, and between the same parallels, are equal to one another 16 (A) If a parallelogram and a triangle stand upon thesarae base, and between the same... | |
 | Augustus De Morgan - 1831 - 108 pages
...two angles of the others, and the interjacent side equal, are equal in all respects. Parallelograms on the same or equal bases, and between the same parallels, are equal. The explanation of this is as follows : the whole proposition is divided into distinct assertions,... | |
 | Robert Gibson - Surveying - 1832 - 290 pages
...ABCD being a parallelogram, AB will be = CD, and AD=BC. THEOREM XIII. PL. l.fig.3l All parallelograms on the same or equal bases and between the same parallels are equal to one another ; that is, if BD=GH, and the lines BH and AF are parallel, then &e parallelogram ABDC—BDFE—EFHG.... | |
 | Mathematics - 1835 - 684 pages
...altitude . cor. 16 • From two Greek words, signifying " along one another.* (i) Parallelograms upon the same or equal bases, and between the same parallels, are equal to one another 16 (/.•) If a parallelogram and a triangle stand upon the same base, and between the... | |
 | Mathematics - 1836 - 530 pages
...angles of the others, and the interjacent side equal, are equal in all respects. /, g Parallelograms on the same or equal bases, and between the same parallels, are equal. The explanation of this is as follows : the whole proposition is divided into distinct assertions,... | |
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XIII. •All parallelograms on the same or equal bases and between the same parallels... 
