| 1867 - 964 páginas
...any point in tho straight line HK, produced both ways indefinitely. Triangles also which stand npon equal bases and between the same parallels are equal to one another. Thus, the triangles LNG, M o F, which „ ._ stand on equal bases, NG, F o, and K between the same... | |
| Robert Gibson - 1795 - 384 páginas
...will be = CD and AD=BC. THEO. Xin. All Parallelograms on the fame or equal Safes and letween the fame Parallels are equal to one another , that is if BD=GH, and the Linet BH and AF parallel, then the Parallelogram ABDC=BDFE=EFHG. For AC (=DB=EF (by Cor. the laft)... | |
| Robert Simson - 1806 - 546 páginas
...is'equal to the triangle DBC. Wherefore, triangles, fee, QE D, PROP. XXXVIII. THEOR. TRIANGLES upon equal bases, and between the same parallels, are equal to one another. Let the triangles ABC, DEF be upon equal bases BC, EF, and between the same parallels BF, AD : the... | |
| John Playfair - 1806 - 320 páginas
...EBCF. Therefore, parallelograms upon the same base, &c. QED PROP. XXXVI. THEOR. PARALLELOGRAMS upon equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and between the same parallels AH, BG; the... | |
| Robert Gibson - 1808 - 482 páginas
...the parallelogram EFGH=BDEF. Wherefore ABDC=BDEF=EFHG. QED " Cor. Hence it is plain that triangles on the same or equal bases and between the same parallels, are equal, seeing (by cor. 2. theo. 12.) they are the halves of their respective parallelograms. THEO. XIV. In... | |
| Robert Gibson - 1811 - 580 páginas
...|»rallelogram E FG H= B DEF. Wherefore ABDC=> BDEF=EFHG. 3. ED Cor. Hence it is plain that triangles on the same or equal bases, and between the same parallels, are equal, seoing (by cor. 2. theo. 1Q.) theyavf the halves of their respective parallelogram • THEO. XIV. PL.... | |
| John Mason Good - 1813 - 714 páginas
...base, and between the same parallels, are equal to one another. Prop. XXXVIII. Theor. Triangles upon equal bases, and between the same parallels, are equal to one another. Prop. XXXIX. Theor. Equal triangles upon the same base, and upon the same side of it, are between the... | |
| Robert Gibson - 1814 - 558 páginas
...the parallelogram EFGH=BDEF. Wherefore ABDC=BDEF=EFHG. QED ч Cor. Hence it is plain that triangles on the same or equal bases, and between the same parallels, are equal, seeing (by cor. 2. theo. 12.) they are the halves of their respective parallelogram. THEO. XIV. PL.... | |
| Euclides - 1816 - 588 páginas
...Therefore parallelograms upon the same base, &c. QED 1 PROP. XXXVI. THEOR. Boot I. PARALLELOGRAMS upon equal bases, and between the same parallels, are equal to one another. •f> K LetABCD,EFGH,be parallelograms upon equal bases BC,FG, and between the same parallels AH, BG;... | |
| Robert Gibson - 1818 - 502 páginas
...prove the parallelogram EFGH=BDEF. Wherefore ABCD=BDEF=EFHG. QED Cor. Hence it is plain that triangles on the same or equal bases and between the same parallels, are equal, seeing (by cor. 2. theo. 12.) they are the halves of their respective parallelograms. THEOREM XIV.... | |
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