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Proposition 55

If an area is contained by a rational straight line and the second binomial, then the side of the area is the irrational straight line which is called a first bimedial.
Let the area ABCD be contained by the rational straight line AB and the second binomial AD. I say that the side of the area AC is a first bimedial straight line.
Since AD is a second binomial straight line, divide it into its terms at E, so that AE is the greater term. Then AE and ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and the lesser term ED is commensurable in length with AB. | X.Def.II.2
Bisect ED at F, and apply to AE the rectangle AG by GE equal to the square on EF and deficient by a square figure. Then AG is commensurable in length with GE. | X.17
Draw GH, EK, and FL through G, E, and F parallel to AB and CD. Construct the square SN equal to the parallelogram AH, and the square NQ equal to GK, and place them so that MN is in a straight line with NO. Then RN is also in a straight line with NP. Complete the square SQ. It is then manifest from what was proved before that MR is a mean proportional between SN and NQ and equals EL, and that is the side of the area AC. It is now to be proved that MO is a first bimedial straight line.
Since AE is commensurable in length with ED, while ED is commensurable with AB, therefore AE is incommensurable with AB. | X.13
Since AG is commensurable with EG, therefore AE is also commensurable with each of the straight lines AG and GE. | X.15
But AE is incommensurable in length with AB, therefore AG and GE are also incommensurable with AB. | X.13
Therefore BA and AG, and BA and GE, are pairs of rational straight lines commensurable in square only, so that each of the rectangles AH and GK is medial. | X.21
Hence, each of the squares SN and NQ is medial. Therefore MN and NO are also medial.
Since AG is commensurable in length with GE, therefore AH is also commensurable with GK, that is, SN is commensurable with NQ, that is, the square on MN with the square on NO. | VI.1 X.11
Since AE is incommensurable in length with ED, while AE is commensurable with AG, and ED is commensurable with EF, therefore AG is incommensurable with EF, so that AH is also incommensurable with EL, that is, SN is incommensurable with MR, that is, PN with NR, that is, MN is incommensurable in length with NO. | X.13 VI.1 X.11
But MN and NO were proved to be both medial and commensurable in square, therefore MN and NO are medial straight lines commensurable in square only. I say next that they also contain a rational rectangle.
Since DE is, by hypothesis, commensurable with each of the straight lines AB and EF, therefore EF is also commensurable with EK. | X.12
And each of them is rational, therefore EL, that is, MR is rational, and MR is the rectangle MN by NO. | X.19
But, if two medial straight lines commensurable in square only and containing a rational rectangle are added together, then the whole is irrational and is called a first bimedial straight line. Therefore MO is a first bimedial straight line. | X.37
Therefore, if an area is contained by a rational straight line and the second binomial, then the side of the area is the irrational straight line which is called a first bimedial.
Q.E.D.

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This proposition is used in X.71.

Book X Introduction - Proposition X.54 - Proposition X.56.

© 1996
D.E.Joyce
Clark University