# Proposition 2

If two straight lines cut one another, then they lie in one plane; and every triangle lies in one plane.

For let the two straight lines AB and CD cut one another at the point E.

I say that AB and CD lie in one plane, and that every triangle lies in one plane.

Take the points F and G at random on EC and EB, join CB and FG, and draw FH and GK across.

I say first that the triangle ECB lies in one plane.

For, if part of the triangle ECB, either FHC or GBK, is in the plane of reference, and the rest in another, then a part also of one of the straight lines EC or EB is in the plane of reference, and a part in another.

XI.1

But, if the part FCBG of the triangle ECB is in the plane of reference, and the rest in another, then a part also of both the straight lines EC and EB is in the plane of reference and a part in another, which was proved absurd.

Therefore the triangle ECB lies in one plane.

XI.1

But, in whatever plane the triangle ECB lies, each of the straight lines EC and EB also lies, and in whatever plane each of the straight lines EC and EB lies, AB and CD also lie.

Therefore the straight lines AB and CD lie in one plane; and every triangle lies in one plane.

Therefore, if two straight lines cut one another, then they lie in one plane; and every triangle lies in one plane.

Q.E.D.

## Guide

The goal of the proof in this proposition is to produce a plane for the two lines AB and CD to lie in. Yet the proof fails to produce any plane at all. Near the beginning is the phrase “the plane of reference” occurs, but there is no reference as no planes have been mentioned. As the two lines AB and CD could be placed anywhere in space, any previously conceived plane would be irrelevant to them.

Postulates of some sort are needed to justify the existence of planes. One could state that three noncollinear points determine a plane. Another might be that there are four noncoplanar points.

#### Use of this proposition

This proposition is used in the proofs of propositions XI.4, XI.6, and XII.17.