 | Euclid - 1845 - 218 σελίδες
...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... | |
 | Euclid, James Thomson - 1845 - 382 σελίδες
...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... | |
 | Euclides - 1845 - 546 σελίδες
...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... | |
 | Euclides - 1846 - 292 σελίδες
...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... | |
 | Euclid, John Playfair - 1846 - 334 σελίδες
...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... | |
 | W. PEASE - 1846 - 86 σελίδες
...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 σελίδες
...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... | |
 | Great Britain. Admiralty - 1846 - 128 σελίδες
...•=• 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... | |
 | London univ - 1846 - 326 σελίδες
...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 σελίδες
...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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TRIANGLES upon the same base, and between the same parallels, are equal to one another. 
