Commentary on Proposition II.13¶
Back to Proposition II.13.
As the text stands, this proposition is unequivocally enunciated of acute-angled triangles; and, as if to obviate any doubt as to whether the restriction was fully intended, the enunciation speaks of the rectangle contained by one of the sides containing the acute angle and the straight line intercepted within by the perpendicular towards the acute angle. On the other hand, it is curious that it speaks of the square on the side subtending the acute angle; and again the setting-out begins “let ABC be an acute-angled triangle having the angle at B acute,” though the last words have no point if all the angles of the triangle are necessarily acute.
It was however very early noticed, not only by Isaacus Monachus, Campanus, Peletarius, Clavius, Commandinus and the rest, but by the Greek scholiast (Heiberg, Vol. V. p. 253), that the relation between the sides of a triangle established by this theorem is true of the side opposite to, and the sides about, an acute angle respectively in any sort of triangle whether acute-angled, right-angled or obtuse-angled. The scholiast tries to explain away the word “acute-angled” in the enunciation: “Since in the definitions he calls acute-angled the triangle which has three acute angles, you must know that he does not mean that here, but calls all triangles acute-angled because all have an acute angle, one at least, if not all. The enunciation therefore is: ‘In any triangle the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle, etc.’”
We may judge too by Heron’s enunciation of his “converse” of the proposition that he would have left the word “acute-angled” out of the enunciation. His converse is: In any triangle in which the square on one of the sides is less than the squares on the other two sides, the angle contained by the latter sides is acute.
If the triangle that we take is a right-angled triangle, and the perpendicular is drawn, not from the right angle, but from the acute angle not referred to in the enunciation, the proposition reduces to I. 47, and this case need not detain us.
The other cases can be proved, like II. 12, after the manner of I. 47.
Let us take first the case where all the angles of the triangle are acute.
As before, if we draw ALP, BMQ, CNR perpendicular to BC, CA, AB and meeting the further sides of the squares on BC, CA, AB in P, Q, R, and if we join KC, AD, we have
the triangles KBC, ABD equal in all respects,
and consequently the rectangles BP, BR equal to one another.
Similarly the rectangles CP, CQ are equal to one another.
Next, by joining BG, CH, we prove in like manner that the rectangles AR, AQ 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 diminished by the rectangles AR, AQ.
But the rectangles AR, AQ are equal, and they are respectively the rectangles contained by BA, AN and by CA, AM.
Therefore the square on BC is less than the squares on BA, AC by twice the rectangle BA, AN or CA, AM.
Next suppose that we have to prove the theorem in the case where the triangle has an obtuse angle at A.
Take B as the acute angle under consideration, so that AC is the side opposite to it.
Now the square on CA is equal to the difference of the rectangles CQ, AQ,
i.e. to the difference between CP and AQ,
i.e. to the difference between the square BE and the sum of the rectangles BP, AQ,
i.e. to the difference between the square BE and the sum of the rectangles BP, AR,
i.e. to the difference between the sum of the squares BE, BH and the sum of the rectangles BP, BR
(since AR is the difference between BR and BH).
But BP, BR are equal, and they are respectively the rectangles CB, BL and AB, BN.
Therefore the square on CA is less than the squares on AB, BC by twice the rectangle CB, BL or AB, BN.
Heron’s proof of his converse proposition (an-Nairīzī, ed. Curtze, p. 110), which is also given by the Greek scholiast above quoted, is of course simple. For let ABC be a triangle in which the square on AC is less than the squares on AB, BC.
Draw BD at right angles to BC and of length equal to BA.
Join DC.
Then, since the angle CBD is right, the square on DC is equal to the squares on DB, BC, i.e. to the squares on AB, BC. [I. 47]
But the square on AC is less than the squares on AB, BC.
Therefore the square on AC is less than the square on DC.
Therefore AC is less than DC.
Hence in the two triangles DBC, ABC the sides about the angles DBC, ABC are respectively equal, but the base DC is greater than the base AC.
Therefore the angle DBC (a right angle) is greater than the angle ABC [I. 25], which latter is therefore acute.
It may be noted, lastly, that II. 12, 13 are supplementary to I. 47 and complete the theory of the relations between the squares on the sides of any triangle, whether right-angled or not.