 | 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... | |
 | Anthony Nesbit - 1847 - 492 σελίδες
...parallelogram ABCD, is equal to the parallelogram DBCE. (Euc. I. 35. Simp. II. 2. Em. HI. 6.) THEOREM VL Let the triangles ABC, DBC be upon the same base BC,...between the same parallels AD, BC ; the triangle ABC is equal to the triangle DBC. (Euc. I. 37. Simp. II. 2. Em. II. 10.) THEOREM VH. Let ABC be a right-angled... | |
 | 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... | |
 | 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... | |
 | 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... | |
 | 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,... | |
 | Thomas Tate (mathematical master.) - 1848 - 284 σελίδες
...ADGK = the parallelogram ADCB; therefore the triangle ADF is also = half the parallelogram ADCB. Cor. Triangles upon the same base and between the same parallels are equal. Application of this Theorem. 1. To show that the rectangle BCGF _—^— A o contains double the surface... | |
 | Euclid, Thomas Tate - 1849 - 120 σελίδες
...diameter BC divides the parallelogram ACDB into two equal parts. QED PROP. XXXV. THEOR. Parallelograms upon the same base and between the same parallels, are equal to one another. Let the parallelograms ABCD, EBCF (see the 2d and 3d figures) be upon the same base BC, and between the same... | |
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Parallelograms upon the same base and between the same parallels, are equal to one another. 
