 | Euclides - 1852 - 152 σελίδες
...annexed, and the Proposition should be proved from it.] AEDH XXXV. 1. BOOK I. PROP. XXXVII. THEOR. Triangles upon the same base, and between the same parallels, are equal to one another. Let the triangles ABC, DEC be upon'the same base BC, and between the same parallels AD, BC : E The... | |
 | Euclides - 1852 - 48 σελίδες
...35.) ; therefore the parallelograms ABCD and EFGH are equal (Ax. 1.). WWD РBОР. XXXVП. ТнЕОB. Triangles upon the same base and between the same parallels are equal. Let the triangles в A c and в D c be on the same base в c, and between the same parallels в c and... | |
 | Royal Military Academy, Woolwich - 1853 - 400 σελίδες
...parallelogram ABCD is equal (1 Ax.) to EFGH. "Wherefore parallelograms, etc. QED 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 - 1853 - 146 σελίδες
...The parallelogram ABCD is equal to EFGH. Wherefore, parallelograms, &c. QED PROP. 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. The triangle... | |
 | Euclides - 1853 - 176 σελίδες
...divides the parallelogram a С db into two equal parts. QED PROPOSITION XXXV. — THEOREM. Parallelograms upon the same base, and between the same parallels, are equal to one another. LET the parallelograms abСd, ebСf (see the second figure) be upon the same base b С, andbetween... | |
 | Thomas Lund - 1854 - 520 σελίδες
...FG is equal to that ratio (170), and join DH. Then DHG is the triangle required. For BDC, BDF, being triangles upon the same base and between the same parallels, are equal to one another (41). Also A.ADE — &ADG; .'.adding to these equals the AABD, it is evident that ADGF = the polygon... | |
 | Robert Potts - 1855 - 1050 σελίδες
...times as great ? The line joining the bisections of two sides of a triangle is parallel to the base. 3. Triangles upon the same base, and between the same parallels are equal to one another. The lines joining the bisections of the sides of any quadrilateral figure, together constitute a parallelogram.... | |
 | Euclides - 1855 - 270 σελίδες
...only which are at the vertices of any two of its opposite angles. PROP. XXXV. THEOREM. Parallelograms upon the same base, and between the same parallels, are equal to one another. Let the parallelograms AС, B Г be upon the same base B С, and and between the same parallels AF,... | |
 | Great Britain. Committee on Education - 1855 - 976 σελίδες
...equal, each to each, namely, those to which the equal sides are opposite, 2. Parallelograms on the same base and between the same parallels are equal to one another. 3. In any triangle, if the square of one of the sides is equal to the squares of the two other sides,... | |
 | Euclides - 1856 - 168 σελίδες
...it is equal to the parallelogram EHGF ; wherefore EHGF is also equal to ABC D. XXXIX.— EUCLID I. 37. Triangles upon the same base and between the same parallels are equal to one another. Let the triangles ABC, DBC (Fig. 31) be upon the same base BC, and between the same parallels AD, BC... | |
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TRIANGLES upon the same base, and between the same parallels, are equal to one another. 
