There is a crucial gap in the text's proof that, in a neutral geometry, *at least* one line can be drawn parallel to a given line through a point not on that line. A proof of this fact is claimed in the first step in the proof that **E5P** is equivalent to **EPP**. We will prove a Proposition which fills this very real gap.

**Definition.** Two lines are said to be perpendicular if their union contains congruent adjacent angles.

**Proposition.** In a neutral geometry, given a line <-P-Q-> and a point A not on this line, there is a line t through A which is perpendicular to <-P-Q->.

**Proof.** Consider the line <-A-P-> (**Postulate 1**). If m*<*APQ=90, we may easily prove that <-A-P-> is the desired perpendicular (Exercise). We assume that m*<*APQ<90, as we may since we may consider any point of the given line to be Q. By **Postulate 12** we construct a ray P--X-> into the opposite side of <-P-Q-> from A, such that m*<*XPQ = m*<*APQ.By **Postulate 2** and **Postulate 3**, we find the point D on P--X-> such that PD=PA. The segment A--D intersects <-P-Q-> at a point B by **Postulate 9**.

We wish to claim that *<*APB = *<*APQ, and that *<*DPB = *<*DPQ. This will prove that m*<*APB = m*<*DPB. As exercises, prove these angle equalities. The key is to establish that Q and B are both interior to *<*APD.

Now by **Postulate 15 (SAS)**, The triangles APB and DPB are congruent and so the angles *<*ABP and *<*DBP are congruent and adjacent. QED

We have now shown that, in a neutral geometry. we may **drop** a perpendicular from a point to a line. It is immediate from **Postulate 12** that we may **raise** a perpendicular from any point *on* a line, and combining these two operations constructs the desired remote parallel. The real geometrical proof, and hence the "neutral flavor", seems to occur in dropping the perpendicular. There is no more reason to accept the existence of the perpendicular without proof than there is to accept the existence of the parallel!