Commentary on Proposition II.9¶
Back to Proposition II.9.
It is noteworthy that, while the first eight propositions of Book II. are proved independently of the Pythagorean theorem I. 47, all the remaining propositions beginning with the 9th are proved by means of it. Also the 9th and 10th propositions mark a new departure in another respect; the method of demonstration by showing in the figures the various rectangles and squares to which the theorems relate is here abandoned.
The 9th and 10th propositions are related to one another in the same way as the 5th and 6th; they really prove the same result which can, as in the earlier case, be comprised in a single enunciation thus: The sum of the squares on the sum and difference of two given straight lines is equal to twice the sum of the squares on the lines.
The semi-algebraical proof of Prop. 9 is that suggested by the remark on the algebraical formulae given at the end of the note on II. 7. It applies with a very slight modification to both II. 9 and II. 10. We will put in brackets the variations belonging to II. 10. The first of the annexed lines is the figure for II. 9 and the second for II. 10.
By II. 4, the square on AD is equal to the squares on AC, CD and twice the rectangle AC, CD.
By II. 7, the squares on CB, CD (CD, CB) are equal to twice the rectangle CB, CD together with the square on BD.
By addition of these equals crosswise, the squares on AD, DB together with twice the rectangle CB, CD are equal to the squares on AC, CD, CB, CD together with twice the rectangle AC, CD.
But AC, CB are equal, and therefore the rectangles AC, CD and CB, CD are equal.
Taking away the equals, we see that the squares on AD, DB are equal to the squares on AC, CD, CB, CD, i.e. to twice the squares on AC, CD.
To show also that the method of geometrical algebra illustrated by II. 1—8 is still effective for the purpose of proving II. 9, 10, we will now prove II. 9 in that manner.
Draw squares on AD, DB respectively as shown in the figure. Measure DH along DE equal to CD, and HL along HE also equal to CD.
Draw HK, LNO parallel to EF, and CNM parallel to DE.
Measure NP along NO equal to CD, and draw PQ parallel to DE.
Now, since AD, CD are respectively equal to DE, DH,
and, since HL is equal to CD, LE is equal to DB.
Similarly, since each of the segments EM, MQ is equal to CD,
Therefore OQ is equal to the square on DB.
We have to prove that the squares on AD, DB are equal to twice the squares on AC, CD.
Now the square on AD includes KM (the square on AC) and CH, HN (that is, twice the square on CD).
Therefore we have to prove that what is left over of the square on AD together with the square on DB is equal to the square on AC.
The parts left over are the rectangles CK and NE, which are equal to KN, PM respectively.
But the latter with the square on DB are equal to the rectangles KN, PM and the square OQ,
Hence the required result follows.