Commentary on Proposition XI.6¶
Back to Proposition XI.6.
If anyone wishes to convince himself of the real necessity for some general agreement as to the order in which propositions in elementary geometry should be taken, let him contemplate the hopeless result of too much independence on the part of editors in the matter of this proposition and its converse, XI. 8.
Legendre adopts a different, and elegant, method of proof; but he applies it to XI. 8, which he gives first, and then deduces XI. 6 from it by reductio ad absurdum. Dr Mehler uses Legendre’s method of proof but applies it to
6, and then gives XI. 8 as a deduction from it. Lardner follows Legendre. Holgate, the editor of a recent American book, gives Euclid’s proof of XI. 6 and deduces XI. 8 by reductio ad absurdum. His countrymen, Schultze and Sevenoak, give XI. 8 first, but put it after, and deduce it from, Eucl. XI. 10; they then give XI. 6, practically as a deduction from XI. 8 by reductio ad absurdum, after a proposition corresponding to Eucl. XI. 11 and 12, and a corollary to the effect that through a given point one and only one perpendicular can be drawn to a given plane.
We will now give the proof of XI. 6 by Legendre’s method (adopted by Smith and Bryant as well as by Mehler).
Let AB, CD be both perpendicular to the same plane MN.
Join BD.
Now, since BD meets AB, CD, both of which are perpendicular to the plane MN in which BD is, the angles ABD, CDB are right angles.
AB, CD will therefore be parallel provided that they are in the same plane.
Through D draw EDF, in the plane MN, at right angles to BD, and make ED equal to DF.
Join BE, BF, AE, AD, AF. Then the triangles BDE, BDF are equal in all respects (by I. 4), so that
BE is equal to BF.
It follows, since the angles ABE, ABF are right, that the triangles ABE, ABF are equal in all respects, and
AE is equal to AF.
[Mehler now argues elegantly thus. If CE, CF be also joined, it is clear that
CE is equal to CF.
Hence each of the four points A, B, C, D is equidistant from the two points E, F. Therefore the points A, B, C, D are in one plane, so that AB, CD are parallel. If, however, we do not use the locus of points equidistant from two fixed points, we proceed as follows.]
The triangles AED, AFD have their sides equal respectively; hence [I. 8] the angles ADE, ADF are equal, so that ED is at right angles to AD.
Thus ED is at right angles to BD, AD, CD; therefore CD is in the plane through AD, BD. [XI. 5]
But AB is in that same plane; [XI. 2] therefore AB, CD are in the same plane.
And the angles ABD, CDB are right; therefore AB, CD are parallel.