 | 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... | |
 | Euclid, John Playfair - 1846 - 334 σελίδες
...adjacent angles of a parallelogram is equal to two right angles. PROP. XXXV. THEOR. Parallelograms upon, the same base and between the same parallels, are equal to one another. > (SEE THE 2d AND 3d FIGURES.) 34 If the sides AD, DF of the parallelograms ABCD, DBCF opposite to... | |
 | 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... | |
 | 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... | |
 | 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... | |
 | Thomas Gaskin - 1847 - 301 σελίδες
...CAMBRIDGE, Nov. 1847. GEOMETRICAL PROBLEMS. ST JOHN'S COLLEGE. DEC. 1830. (No. I.) 1. PARALLELOGRAMS upon the same base and between the same parallels are equal to one another. 2. Of unequal magnitudes,, the greater has a greater ratio to the same than the less. 3. If the diameter... | |
 | Euclides - 1848 - 52 σελίδες
...diameter bisects them, that is, divides them into two equal parts. PROP. XXXV. THEOREM. Parallelograms upon the same base, and between the same parallels, are equal to one another. PROP. XXXVI. THEOREM. PROP. XXXVII. THEOREM. Triangles upon the same base anti between the same parallels,... | |
 | Great Britain. Council on Education - 1848 - 596 σελίδες
...point out how the construction fails when that condition is not fulfilled. 2. Prove that parallelograms upon the same base and between the same parallels are equal to one another. Shew hence that the area of a parallelogram is properly measured by the product of the numbers that... | |
 | J. Goodall, W. Hammond - 1848 - 390 σελίδες
...point out how the construction fails when that condition is not fulfilled. 2. Prove that parallelograms upon the same base and between the same parallels are equal to one another. Show hence that the area of a parallelogram is properly measured by the product of the numbers that... | |
 | Great Britain. Committee on Education - 1848 - 606 σελίδες
...point out how the construction fails when that condition is not fulfilled. 2. Prove that parallelograms upon the same base and between the same parallels are equal to one another. Shew hence that the area of a parallelogram is properly measured by the product of the numbers that... | |
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Parallelograms upon the same base and between the same parallels, are equal to one another. 
