Commentary on Proposition XI.23¶
Back to Proposition XI.23.
The whole difficulty in this proposition is the proof of a fact which makes the construction possible, viz. the fact that, if LMN be a triangle with sides respectively equal to the bases of the isosceles triangles which have the given angles as vertical angles and the equal sides all of the same length, then one of these equal sides, as AB, is greater than the radius LO of the circle circumscribing the triangle LMN.
Assuming that AB is greater than LO, we have only to draw from O a perpendicular OR to the plane of the triangle LMN, to make OR of such a length that the sum of the squares on LO, OR is equal to the square on AB, and to join RL, RM, RN. (The manner of finding OR such that the square on it is equal to the difference between the squares on AB and LO is shown in the Lemma at the end of the text of the proposition. We have already had the same construction in the Lemma after x. 13.)
Then clearly RL, RM, RN are equal to AB and to one another [I. 4 and I. 47].
Therefore the triangles LRM, MRN, NRL have their three sides respectively equal to those of the triangles ABC, DEF, GHK respectively.
Hence their vertical angles are equal to the three given angles respectively; and the required solid angle is constructed.
We return now to the proposition to be proved as a preliminary to the construction, viz. that, in the figures, AB is greater than LO.
It will be observed that Euclid, as his manner is, proves it for one case only, that, namely, in which O, the centre of the circle circumscribing the triangle LMN, falls within the triangle, leaving the other cases for the reader to prove. As usual, however, the two other cases are found in the Greek text, after the formal conclusion of the proposition, as above, ending with the words ὅπερ ἔδει ποιῆσαι. This position for the proofs itself suggests that they are not Euclid’s but are interpolated; and this is rendered certain by the fact that words distinguishing three cases at the point where the centre O of the circumscribing circle is found, “It [the centre] will then be either within the triangle LMN or on one of its sides or without. First let it be within,” are found in the MSS. B and V only and are manifestly interpolated. Nevertheless the additional two cases must have been inserted very early, as they are found in all the best MSS.
In order to give a clear view of the proof of all three cases as given in the text, we will reproduce all three (Euclid’s as well as the others) with abbreviations to make them catch the eye better.
In all three cases the proof is by reductio ad absurdum, and it is proved first that AB cannot be equal to LO, and secondly that AB cannot be less than LO.
Case I.¶
Suppose, if possible, that AB $=$ LO. Then AB, BC are respectively equal to LO, OM; and AC $=$ LM (by construction). Therefore ∠ABC $=$ ∠LOM. Similarly ∠DEF $=$ ∠MON, ∠GHK $=$ ∠NOL, Adding, we have ∠ABC $+$ ∠DEF $+$ ∠GHK $=$ ∠LOM $+$ ∠MON $+$ ∠NOL $=$ four right angles: which contradicts the hypothesis. Therefore AB $\ne$ LO.
Suppose that AB $<$ LO. Make OP, OQ (measured along OL, OM) each equal to AB. Thus, OL, OM being equal also, it follows that PQ is $\parallel$ to LM. Hence LM $:$ PQ $=$ LO $:$ OP; and, since LO $>$ OP, LM, i.e. AC, $>$ PQ. Thus, in $\Delta\text{s}$ POQ, ABC, two sides are equal to two sides, and base AC $>$ base PQ; therefore ∠ABC $>$ ∠POQ, i.e. ∠LOM. Similarly ∠DEF $>$ ∠MON, ∠GHK $>$ ∠NOL, and it follows by addition that ∠ABC $+$ ∠DEF $+$ ∠GHK $>$ (four right angles): which again contradicts the hypothesis.
Case II.¶
Suppose, if possible, that AB $=$ LO. Then (AB $+$ BC), or (DE $+$ EF) $=$ MO $+$ OL $=$ MN $=$ DF; which contradicts the hypothesis.
The supposition that AB $<$ LO is even more impossible; for in this case it would result that DE $+$ EF $<$ DF.
Case III.¶
Suppose, if possible, that AB $=$ LO. Then, in the triangles ABC, LOM, two sides AB, BC are respectively equal to two sides LO, OM, and the bases AC, LM are equal; therefore ∠ABC $=$ ∠LOM. Similarly ∠GHK $=$ ∠NOL. Therefore, by addition, ∠MON $=$ ∠ABC $+$ ∠GHK $>$ ∠DEF (by hypothesis). But, in the triangles DEF, MON, which are equal in all respects, ∠MON $=$ ∠DEF. But it was proved that ∠MON $>$ ∠DEF: which is impossible.
Suppose, if possible, that AB $<$ LO. Along OL, OM measure OP, OQ each equal to AB.
Then LM, PQ are parallel, and LM $:$ PQ $=$ LO $:$ OP, whence, since LO $>$ OP, LM, or AC, $>$ PQ. Thus, in the triangles ABC, POQ, ∠ABC $>$ ∠POQ, i.e. ∠LOM. Similarly, by taking OR along ON equal to AB, we prove that ∠GHK $>$ ∠LON. Now, at O, make ∠POS equal to ∠ABC, and ∠POT equal to ∠GHK. Make OS, OT each equal to OP, and join ST, SP, TP. Then, in the equal triangles ABC, POS, AC $=$ PS, so that LM $=$ PS. Similarly LN $=$ PT. Therefore in the triangles MLN, SPT, since ∠MLN $>$ ∠SPT [this is assumed, but should have been explained], MN $>$ ST, or DF $>$ ST. Lastly, in $\Delta\text{s}$ DEF, SOT, which have two sides equal to two sides, since DF $>$ ST, ∠DEF $>$ ∠SOT $>$ ∠ABC $+$ ∠GHK (by construction): which contradicts the hypothesis.
Simson gives rather different proofs for all three cases; but the essence of them can be put, I think, a little more shortly than in his text, as well as more clearly
Case I. (O within △LMN.)¶
Let AB be, if possible, equal to LO. Then the $\Delta\text{s}$ ABC, DEF, GHK must be identically equal to the $\Delta\text{s}$ LOM, MON, NOL respectively.
Therefore the vertical angles at O in the latter triangles are equal respectively to the angles at B, E, H. The latter are therefore together equal to four right angles: which is impossible.
If AB be less than LO, construct on the bases LM, MN, NL triangles with vertices P, Q, R and identically equal to the $\Delta\text{s}$ ABC, DEF, GHK respectively.
Then P, Q, R will fall within the respective angles at O, since PL = PM and < LO, and similarly in the other cases. Thus [I. 21] the angles at P, Q, R are respectively greater than the angles at O in which they lie. Therefore the sum of the angles at P, Q, R, i.e. the sum of the angles at B, E, H, is greater than four right angles: which again contradicts the hypothesis.
Case II. (O lying on MN.)¶
In this case, whether (1) AB = LO, or (2) AB < LO, a triangle cannot be formed with MN as base and each of the other sides equal to AB. In other words, the triangle DEF either reduces to a straight line or is impossible.
Case III. (O lying outside the △LMN.)¶
Suppose, if possible, that AB = LO.
Then the triangles LOM, MON, NOL are identically equal to the triangles ABC, DEF, GHK. Since ∠LOM + ∠LON = ∠MON, ∠ABC + ∠GHK = ∠DEF: which contradicts the hypothesis.
Suppose that AB < OL.
Draw, as before, on LM, MN, NL as bases triangles with vertices P, Q, R and identically equal to the $\Delta\text{s}$ ABC, DEF, GHK. Next, at N on the straight line NR, make ∠RNS equal to the angle PLM, cut off NS equal to LM and join RS, LS. Then △NRS is identically equal to △LPM or △ABC. Now (∠LNR + ∠RNS) < (∠NLO + ∠OLM), that is, ∠LNS < ∠NLM. Thus, in $\Delta\text{s}$ LNS, NLM, two sides are equal to two sides, and the included angle in the former is less than the included angle in the other. Therefore LS < MN.
Hence, in the triangles MQN, LRS, two sides are equal to two sides, and MN > LS. Therefore ∠MQN > ∠LRS > (∠LRN + ∠SRN) > (∠LRN + ∠LPM). That is, ∠DEF > (∠GHK + ∠ABC): which is impossible.