All 13 books of Euclid's Elements, the basis of mathematics.
Proposition 18 If there are two unequal straight-lines, and a (rect angle) equal to the fourth part of the (square) on the , lesser, falling short by a square gure, is applied to the [], greater, and divides it into (parts which are) incom . mensurable [in length], then the square on the greater , will be larger than the (square on the) lesser by the (square) on (some straight-line) incommensurable (in length) with the greater. And if the square on the , []. , , , greater is larger than the (square on the) lesser by the (square) on (some straight-line) incommensurable (in [] length) with the greater, and a (rectangle) equal to the , fourth (part) of the (square) on the lesser, falling short by , , a square gure, is applied to the greater, then it divides it .
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, into (parts which are) incommensurable [in length]. Let A and BC be two unequal straight-lines, of which (let) BC (be) the greater. And let a (rectangle) equal to the fourth [part] of the (square) on the lesser, A, falling short by a square gure, have been applied to BC. And let it be the (rectangle contained) by BDC. And let BD be incommensurable in length with DC. I say that that the square on BC is greater than the (square on) A by the (square) on (some straight-line) incommensurable (in length) with (BC). 299 x ELEMENTS BOOK 10
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A E For, similarly, by the same construction as before, we can show that the square on BC is greater than the , . (square on) A by the (square) on F D. [Therefore] it [], . , must be shown that BC is incommensurable in length . with DF . For since BD is incommensurable in length , with DC, BC is thus also incommensurable in length , . with CD [Prop. 10.16]. But, DC is commensurable (in length) with the sum of BF and DC [Prop. 10.6]. And, . thus, BC is incommensurable (in length) with the sum of BF and DC [Prop. 10.13]. Hence, BC is also incommen . surable in length with the remainder F D [Prop. 10.16]. , And the square on BC is greater than the (square on) , A by the (square) on F D. Thus, the square on BC is , . , greater than the (square on) A by the (square) on (some . B F D C
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So, again, let the square on BC be greater than the . (square on) A by the (square) on (some straight-line) incommensurable (in length) with (BC). And let a (rect . angle) equal to the fourth [part] of the (square) on A, falling short by a square gure, have been applied to BC. , . , And let it be the (rectangle contained) by BD and DC. It must be shown that BD is incommensurable in length with DC. . ,
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