Commentary on Proposition II.12¶
Back to Proposition II.12.
Since in this proposition and the next we have to do with the squares on the sides of triangles, the particular form of graphic representation of areas which we have had in Book II. up to this point does not help us to visualise the results of the propositions in the same way, and only two lines of proof are possible, (1) by means of the results of certain earlier propositions in Book II. combined with the result of I. 47 and (2) by means of the procedure in Euclid’s proof of I. 47 itself. The alternative proofs of II. 12, 13 after the manner of Euclid’s proof of I. 47 are therefore alone worth giving.
These proofs appear in certain modern text-books (e.g. Mehler, Henrici and Treutlein, H. M. Taylor, Smith and Bryant). Smith and Bryant are not correct in saying (p. 142) that they cannot be traced further back than Lardner’s Euclid (1828); they are to be found in Gregory of St. Vincent’s work (published in 1647) Opus geometricum quadraturae circuli et sectionum coni, Book I. Pt. 2, Props. 44, 45 (pp. 31, 32).
To prove II. 12, take an obtuse-angled triangle ABC in which the angle at A is the obtuse angle Describe squares on BC, CA, AB, as BCED, CAGF, ABKH.
Draw AL, BM, CN, perpendicular to BC, CA, AB (produced if necessary), and produce them to meet the further sides of the squares on them in P, Q, R respectively.
Join AD, CK.
Then, as in I. 47, the triangles KBC, ABD are equal in all respects;
therefore their doubles, the parallelograms in the same parallels respectively, are equal;
that is, the rectangle BP is equal to the rectangle BR.
Similarly the rectangle CP is equal to the rectangle CQ.
Also, if BG, CH be joined, we see that
the triangles BAG, HAC are equal in all respects;
therefore their doubles, the rectangles AQ, AR, are equal.
Now the square on BC is equal to the sum of the rectangles BP, CP,
i.e. to the sum of the rectangles BR, CQ,
i.e. to the sum of the squares BH, CG and the rectangles AR, AQ.
But the rectangles AR, AQ are equal, and they are respectively the rectangle contained by BA, AN and the rectangle contained by CA, AM.
Therefore the square on BC is equal to the squares on BA, AC together with twice the rectangle BA, AN or CA, AM.
Incidentally this proof shows that the rectangle BA, AN is equal to the rectangle CA, AM: a result which will be seen later on to be a particular case of the theorem in III. 35.
Heron (in an-Nairīzī, ed. Curtze, p. 109) gives a “converse” of II. 12 related to it as I. 48 is related to I. 47.
In any triangle, if the square on one of the sides is greater than the squares on the other two sides, the angle contained by the latter is obtuse.
Let ABC be a triangle such that the square on BC is greater than the squares on BA, AC.
Draw AD at right angles to AC and of length equal to AB.
Join DC.
Then, since DAC is a right angle, the square on DC is equal to the squares on DA, AC, [I. 47]
i.e. to the squares on BA, AC.
But the square on BC is greater than the squares on BA, AC; therefore the square on BC is greater than the square on DC.
Therefore BC is greater than DC.
Thus, in the triangles BAC, DAC, the two sides BA, AC are equal to the two sides DA, AC respectively,
but the base BC is greater than the base DC.
Therefore the angle BAC is greater than the angle DAC; [I. 25]
that is, the angle BAC is obtuse.