Part 7

# Propositions

by Euclid

## Proposition 25

The rectangle contained by medial straight-lines (which are) commensurable in square only is either ra- tional or medial.

A G C D B H M K N E L For let the rectangle AC be contained by the medial straight-lines AB and BC (which are) commensurable in square only. I say that AC is either rational or medial. For let the squares AD and BE have been described on (the straight-lines) AB and BC (respectively). AD and BE are thus each medial. And let the rational (straight-line) FG be laid out. And let the rectangular parallelogram GH, equal to AD, have been applied to FG, producing FH as breadth. And let the rectangular parallelogram MK, equal to AC, have been applied to HM, producing HK as breadth. And, finally, let NL, equal to BE, have similarly been applied to KN, pro- ducing KL as breadth. Thus, FH, HK, and KL are in a straight-line. Therefore, since AD and BE are each medial, and AD is equal to GH, and BE to NL, GH and NL (are) thus each also medial. And they are ap- plied to the rational (straight-line) FG. FH and KL are thus each rational, and incommensurable in length with FG [Prop. 10.22]. And since AD is commensurable with BE, GH is thus also commensurable with NL. And as

GH is to NL, so FH (is) to KL [Prop. 6.1]. Thus, FH is commensurable in length with KL [Prop. 10.11]. Thus, FH and KL are rational (straight-lines which are) com- mensurable in length. Thus, the (rectangle contained) by FH and KL is rational [Prop. 10.19]. And since DB is equal to BA, and OB to BC, thus as DB is to BC, so AB (is) to BO. But, as DB (is) to BC, so DA (is) to AC [Props. 6.1]. And as AB (is) to BO, so AC (is) to CO [Prop. 6.1]. Thus, as DA is to AC, so AC (is) to CO. And AD is equal to GH, and AC to MK, and CO to NL. Thus, as GH is to MK, so MK (is) to NL. Thus, also, as FH is to HK, so HK (is) to KL [Props. 6.1, 5.11]. Thus, the (rectangle contained) by FH and KL is equal to the (square) on HK [Prop. 6.17]. And the (rectangle contained) by FH and KL (is) rational. Thus, the (square) on HK is also rational. Thus, HK is ratio- nal. And if it is commensurable in length with FG then HN is rational [Prop. 10.19]. And if it is incommensu- rable in length with FG then KH and HM are rational (straight-lines which are) commensurable in square only: thus, HN is medial [Prop. 10.21]. Thus, HN is either ra- tional or medial. And HN (is) equal to AC. Thus, AC is either rational or medial. Thus, the… by medial straight-lines (which are) com- mensurable in square only, and so on…. Proposition 26 A medial (area) does not exceed a medial (area) by a rational (area). A E D C -B K G H For, if possible, let the medial (area) AB exceed the medial (area) AC by the rational (area) DB. And let the rational (straight-line) EF be laid down. And let the rectangular parallelogram FH, equal to AB, have been applied to to EF, producing EH as breadth. And let FG, equal to AC, have been cut off (from FH). Thus, the remainder BD is equal to the remainder KH. And DB is rational. Thus, KH is also rational. Therefore, since AB and AC are each medial, and AB is equal to FH, and AC to FG, FH and FG are thus each also medial.

And they are applied to the rational (straight-line) EF. Thus, HE and EG are each rational, and incommensu rable in length with EF [Prop. 10.22]. And since DB is rational, and is equal to KH, KH is thus also ratio- nal. And (KH) is applied to the rational (straight-line) EF. GH is thus rational, and commensurable in length with EF [Prop. 10.20]. But, EG is also rational, and in- commensurable in length with EF. Thus, EG is incom- mensurable in length with GH [Prop. 10.13]. And as EG is to GH, so the (square) on EG (is) to the (rectan- gle contained) by EG and GH [Prop. 10.13 lem.]. Thus, the (square) on EG is incommensurable with the (rect- angle contained) by EG and GH [Prop. 10.11]. But, the (sum of the) squares on EG and GH is commensurable with the (square) on EG. For (EG and GH are) both rational. And twice the (rectangle contained) by EG and GH is commensurable with the (rectangle contained) by EG and GH [Prop. 10.6]. For (the former) is double the latter. Thus, the (sum of the squares) on EG and GH is incommensurable with twice the (rectangle con- tained) by EG and GH [Prop. 10.13]. And thus the sum of the (squares) on EG and GH plus twice the (rectan- gle contained) by EG and GH, that is the (square) on EH [Prop. 2.4], is incommensurable with the (sum of the squares) on EG and GH [Prop. 10.16]. And the (sum of the squares) on EG and GH (is) rational. Thus, the (square) on EH is irrational [Def. 10.4]. Thus, EH is irrational [Def. 10.4]. But, (it is) also rational. The very thing is impossible. Thus, a medial (area) does not exceed a medial (area) by a rational (area). (Which is) the very thing it was required to show.

## Proposition 27

To find (two) medial (straight-lines), containing a ra- tional (area), (which are) commensurable in square only. A B C D Let the two rational (straight-lines) A and B, (which are) commensurable in square only, be laid down. And let C-the mean proportional (straight-line) to A and B-

have been taken [Prop. 6.13]. And let it be contrived that as A (is) to B, so C (is) to D [Prop. 6.12]. And since the rational (straight-lines) A and B are commensurable in square only, the (rectangle con- tained) by A and B-that is to say, the (square) on C [Prop. 6.17]-is thus medial [Prop 10.21]. Thus, C is medial [Prop. 10.21]. And since as A is to B, [so] C (is) to D, and A and B [are] commensurable in square only, C and D are thus also commensurable in square only [Prop. 10.11]. And C is medial. Thus, D is also medial [Prop. 10.23]. Thus, C and D are medial (straight-lines which are) commensurable in square only. I say that they also contain a rational (area). For since as A is to B, so C (is) to D, thus, alternately, as A is to C, so B (is) to D [Prop. 5.16]. But, as A (is) to C, (so) C (is) to B. And thus as C (is) to B, so B (is) to D [Prop. 5.11]. Thus, the (rectangle contained) by C and D is equal to the (square) on B [Prop. 6.17]. And the (square) on B (is) rational. Thus, the (rectangle contained) by C and D [is] also rational. Thus, (two) medial (straight-lines, C and D), con- taining a rational (area), (which are) commensurable in square only, have been found. (Which is) the very thing it was required to show. re the length of B is k13 times that of A.

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