| Robert Gibson - 1806 - 452 sider
...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 - 1811 - 508 sider
...BDEF ; in like manner we may prove the |»rallelogram 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 - 1814 - 508 sider
...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 - 1818 - 478 sider
...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 - 1821 - 544 sider
...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 - 1830 - 272 sider
...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
...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 - 1832 - 348 sider
...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.... | |
| 1835
...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... | |
| 1836
...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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