If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part. The Elements of Euclid - Side 41av Euclid - 1838 - 416 siderUten tilgangsbegrensning - Om denne boken
| Stewart W. and co - 1884
...together with the rectangle AB, BC, is equal to the square of AB. III. — If a straight line be divided **into any two parts, the rectangle contained by the...by the two parts, together with the square of the** aforesaid part. Let AB be divided into any two parts in the point C; then the rectangle AB, BC, is... | |
| Euclides - 1884
...AB ; NOTE. — The enunciation of this proposition usually given is : If a straight line be divided **into any two parts, the rectangle contained by the...contained by the two parts together with the square** on the aforesaid part. That is, in reference to the figure, an expression which can be easily derived... | |
| Cambridge univ, exam. papers - 1884
...respectively. Prove that the triangle DBG is double the triangle DEC. •i. If a straight line be divided **into any two parts, the rectangle contained by the...contained by the two parts, together with the square** on the aforesaid part. 5. Describe a square that shall be equal to a given rectilineal figure. On the... | |
| Oxford univ, local exams - 1885
...right angles. Enunciate and prove the corollaries of this proposition. 3. If a straight line be divided **into any two parts, the rectangle contained by the...contained by the two parts, together with the square** on the aforesaid part. 4. The angles in the same segment of a circle are equal to one another. 5. About... | |
| Dalhousie University - 1885
...lines towards the same parts, are themselves equal and parallel. 10. If a straight line be divided **into any two parts, the rectangle contained by the whole and one of the parts is equal to the** square of that part together with the rectangle of the two paits. GEOMETRY. (EXHIBITIONS AND BURSARIES.)... | |
| GEORGE BRUCE HALSTED - 1885
...Therefore 102 293. If c = a, then a(b + c) = a(b + a) = aa + a*. a*c Therefore If a sect be divided **into any two parts, the rectangle contained by the whole and one of the parts is** equivalent to the rectangle contained by the two parts, together with the square on the aforesaid part.... | |
| George Bruce Halsted - 1886 - 366 sider
...IO2 293. If c = a, then a(b + e) = a(6 + a) = ab + aa = ab + of. ac Therefore If a sect be divided **into any two parts, the rectangle contained by the whole and one of the parts is** equivalent to the rectangle contained by the two parts, together with the square on the aforesaid part.... | |
| Dalhousie University - 1888
...parallel. 3. If a straight line be diyided into any two parts, the rectangle contained by the whole line **and one of the parts, is equal to the rectangle contained...by the two parts, together with the square of the** aforesaid part. 4. Prove, either by a diagram or in any other way, that if a straight line be bisected... | |
| E. J. Brooksmith - 1889
...parallel to the other diagonal, intersect in E. Prove that BE= CD. 3. If a straight line be divided **into any two parts, the rectangle contained by the...contained by the two parts, together with the square** on the aforesaid part. The diagonal AC of a square ABCD is produced to E, so that CE=BC. Prove that... | |
| Nathan Fellowes Dupuis - 1889 - 294 sider
...the construction of 183° solves the problem, " To divide a given segment so that the rectangle on **the whole and one of the parts is equal to the rectangle** on the other part and the segment which is the sum of the whole and the first part." r i. Construct... | |
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