By Raymond Clare Archibald

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**Extra info for Euclid's Book on Divisions of Figures**

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48 EUCLID'S BOOK ON DIVISIONS OF FIGURES III [37-38 In the first case join be and through k draw kl\ eb. el [then el is the required bisector]. Join If the point e be between a and k [a similar construction with the line through k parallel to be, and meeting be in my leads to the solution by the line em\. Were e at the middle of a side such as ab, draw dz\\ab and bisect dz in i. Join ei, ci and ec. Through i draw it || ec. ]. If dfe were to fall outside the quadrilateral, draw from c the parallel to ba; and so on.

X. 16 l e m m a ] = AB. AC105. F i n d , by 11. 14, the side, b, of a square equal in area to the rectangle AB. A Q then t h e p r o b l e m is exactly equivalent to that of which a simple solution was given by Simson 1 0 6 : thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect. T h e Proposition 18 of Euclid under consideration is a particular case of this problem and as the fragment of the text and Woepcke's note (note 104) are contained in it, doubt may well be entertained as to whether Euclid gave any construction in his book On Divisions.

Paris, i860, pp. ) The first part of this lemma is practically equivalent to either (1) [vi. edidh of their sides ; or (2) the first part of Prop. 70 of the Data {Euclidis H. Menge, Lipsiae, 1896, p. e. one side in one to one side in the other], the parallelograms themselves will also have a given ratio to one another. Cf H E A T H , Thirteen Books of Euclid's Ele7nents, 11, 250. The proposition is stated in another way by Pappus 8 5 (p. 928) who proves that is to an equiangular parallelogram as the rectangle contained by a parallelogram the adjacent sides of the first is to the rectangle contained by the adjacent sides of the second.