| Euclides, James Hamblin Smith - 1883
...the line, PROPOSITION III. THEOREM. // a straight line be divided into any two parts, the rectangk **contained by the whole and one of the parts is equal to the** netangle contained by the two parts together with the square on the aforesaid part. J) Let the st.... | |
| 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... | |
| University of Cambridge - 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... | |
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