| Euclid - 1845 - 218 páginas
...Wherefore parallelograms, &c. QED 31. 1. 4 Def. 34. ] t 35. 1. 34. 1. TAx: PROPOSITION XXXVII. THEOR. — Triangles upon the same base, and between the same parallels, are equal to one another. Let the triangles ABC, DBC be upon the same base BC, and between the same parallels AD, BC : the triangle... | |
| Euclides - 1845 - 546 páginas
...parallelogram EFGH. (ax. 1.) Therefore parallelograms upon equal, &c. QED PROPOSITION XXXVII. THEOREM. Triangles upon the same base, and between the same parallels, are equal to one another. Let the triangles ABC, DBC be upon the same base BC, and between the same parallels AD, BC. Then the... | |
| Euclid, James Thomson - 1845 - 382 páginas
...parallelogram AC is equal (I. ax. 1) to EG. Wherefore parallelograms, &c. PROP. XXXVII. TnEOR.t — Triangles upon the same base, and between the same parallels, are equal to one another. * lt will appear, from this proposition, that the perimeters of two equal parallelograms on the same... | |
| Euclid, John Playfair - 1846 - 334 páginas
...and joined towards the same parts by the straight lines BE, CH : But __ H B F G PROP. XXXVII. THEOR. Triangles upon the same base, and between the same parallels, are equal to one another. Let the triangles ABC, DBC be upon the same base BC, and between the same parallels, AD, BC : The triangle... | |
| Euclides - 1846 - 292 páginas
...also the parallelogram ABCD is equal to EFGH. 'Wherefore, Parallelograms %c. QED PROP. XXXVII. THEOR. Triangles upon the same base, and between the same parallels, are equal to one another. Let the triangles ABC, DBC be upon the same base BC, and between the same parallels AD, BC : the triangle... | |
| W. PEASE - 1846 - 86 páginas
...form the isosceles triangle required. The reason of this is (Prob. XXXVII. Bk. I. Euclid,) because triangles upon the same base, and between the same parallels, are equal to one another : ie the triangles ACB and AE B, being upon the base, AB, to which the line EC is parallel, therefore... | |
| Great Britain. Admiralty - 1846 - 128 páginas
...•=• parts. Wherefore the opposite sides and angles, &c. PROP. XXXIV. THEOR. 3s. lEu. Parallelograms upon the same base, and between the same parallels, are equal to one another. PROP. XXXIV. FADE F \ . If the sides AD, DF, of the 1=1™ ABCD, DBCF, opp. to BC the base, be terminated... | |
| Great Britain. Admiralty - 1846 - 128 páginas
...two = parts. Wherefore the opposite sides and angles, &c. PROP. XXXIV. THEOR. 35. lEu. Parallelograms upon the same base, and be-tween the same parallels, are equal to one another. FA DEFAEDF \ . If the sides AD, DF, of the / — 7"" ABCD, DBCF, opp. to RC the base, be terminated... | |
| London univ - 1846 - 326 páginas
...exterior angles of any rectilineal figure are together equal to four right angles. 6. Parallelograms upon the same base and between the same parallels are equal to one another. 7. Show that the complements of the parallelograms which are about the diameter of any .parallelogram... | |
| Euclides - 1847 - 128 páginas
...KLNC (Ax.2) = Dm BMNC. Wherefore the sum of the areas &c. — QED PEOP. XXXVII. THEOR. GEN. ENUN. — Triangles upon the same base, and between the same parallels, are equal to one another. PART. ENUN. — Let the A ABC, DBC be upon the same base BC, and between the same || s AD, BC ; then... | |
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