Commentary on Proposition XI.3

Back to Proposition XI.3.

I think Simson is right in objecting to the words after “which is absurd,” to the effect that DEB, DFB are not straight lines, and that neither can there be any other straight line joined from D to B except DB, as being unnecessary. It is right to conclude at once from the absurdity that BD cannot but be a straight line.

Legendre makes his proof depend on Prop. 2. “For, if, among the points common to the two planes, three should be found which are not in a straight line, the two planes in question, each passing through three points, would only amount to one and the same plane.” [This of course assumes that three points determine one and only one plane, which, strictly speaking, involves more than Prop. 2 itself, as shown in the last note.]

A favourite proposition in modern text-books is the following. The proof seems to be due to von Staudt (Killing, Grundlagen der Geometrie, Vol. II. p. 43).

If two planes meet in a point, they meet in a straight line.

Let ABC, ADE be two given planes meeting at A.

Take any points B, C lying on the plane ABC, and not on the plane ADE but on the same side of it.

Join AB, AC, and produce BA to F.

Join CF.

Then, since B, F are on opposite sides of the plane ADE, C, F are also on opposite sides of it.
Therefore CF must meet the plane ADE in some point, say G.
Then, since A, G are both in each of the planes ABC, ADE, the straight line AG is in both planes. [Simson's def.]
B C D A G E F

This is also the place to insert the proposition that, If three planes intersect two and two, their lines of intersection either meet in a point or are parallel two and two.

Let there be three planes intersecting in the straight lines AB, CD, EF.

B A D C F E B A D C F E

Now AB, EF are in a plane; therefore they either meet in a point or are parallel.

  1. Let them meet in O.

Then O, being a point in AB, lies in the plane AD, and, being also a point in EF, lies also in the plane ED.

Therefore O, being common to the planes AD, DE, must lie on CD, the line of their intersection; i.e. CD, if produced, passes through O.

  1. Let AB, EF not meet, but let them be parallel.

Then CD cannot meet AB; for, if it did, it must necessarily meet EF, by the first case.

Therefore CD, AB, being in one plane, are parallel.
Similarly CD, EF are parallel.