Prove equivalence to the Euclidean Parallel Postulate

Show that this statement (P):

The opposite sides of a parallelogram are congruent

is equivalent to the H.E.P.P (Q):

For every line $l$ and every point $p$ not lying on $l$ there is at most one line $m$ through $p$ such that $m\parallel l$

My first idea was to try to show that $Q\implies P$. I tried to do this by using an RAA proof.

1) I began by assuming the H.E.P.P and constructing a transversal to lines $m$ and $l$.

2) Then we can take a point that's not on the transversal that I already constructed, but still on one of the lines, $m$ or $l$, (Without loss of Generality, it doesn't matter which line).

3) Then draw a line through that point that is parallel to my previously constructed transversal, which gives me a parallelogram.

4) Here's where it gets cloudier: I believe my next step is to start talking about the angles created to prove that the opp. angles are congruent, but I'm not sure.

I'm going to work more on $P \implies Q$. I'll edit when I have more on that. I believe that way will be more straightforward. Thanks in advance!

EDIT:

So I believe I figured it out. (starting at point 4))

4) Draw another transversal through the corners of the previously drawn parallelogram.

5) Now we can show that triangle is ABD is congruent to triangle CDB by ASA

6) Therefore side $\overline {AB}=\overline {DC}$ and $\overline {AD}=\overline {CB}$ because of the congruent triangles.

7) Therefore the parallelogram has opp. sides that are congruent.

For showing $P \implies Q$ we just work backwards.

HINT ONLY

I am going to give only a sketchy proof of one direction. Doing such an argumentation the right way would be very lengthy. The moral of my sketch is that the axiom of parallelism (whichever version we consider) cannot be used without referring to the other axioms of Euclidean geometry.

Rather than this wording:

"For every line $l$ and every point $p$ not lying on $l$ there is at most one line $m$ through $p$ such that $m\parallel l$."

I would use the original Euclidean axiom:

If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.

Then I would define parallelism as follows:

Distinct straight lines lying in the same plane are called parallel if they don't intersect.

First, I would have to prove that for any given straight line and any given point lying in the same plane (the point not lying on the straight line) there is exactly one straight line through the given point which does not intersect the given line.

Let the straight line $a$ and the point $P$ be given as shown below.

Select a point $P'$ on $a$ and construct the line $PP'$. (Here we immediately use the axiom that two distinct points determine one unique straight line.) Now, copy the angle $\alpha$ at $P'$ to $P$ as shown. By this the "parallel" line is given. Note that when talking about the possibility of copying an angle we use a series of axioms determining the nature of congruence. $a$ and the constructed "parallel" straight don't intersect. If they intersected on one side of $PP'$ then they would intersect on the other side as well. Just mirror the figure over the line $PP'$ and think of the fact that $\alpha +\beta$ equals exactly two right angles. If our lines intersected in two points then they would not be distinct straights. (Two points determine one unique lines.)

Consider the parallelogram $ABCD$ below. Drop perpendiculars from $A$ and $D$ to $BC$.

Recall that the opposite sides of the parallelogram are parallel. A simple argumentation as the one already used above I can deduce that the corresponding angles of the generated triangles, $ABR$ and $DCR$ equal. For example, to show that $\alpha=\alpha'$ think of $AB$ and $DC$ as parallel lines and $BC$ as a transversal to them: $\alpha+\gamma'$ has to equal two right angles. Also, $\gamma+\alpha=\gamma'+\alpha'$ equal two right angles... So our two triangles are congruent. As a result $AB=DC$. (For $BC=AD$ we can use the same argumentation.)

Again I've been using the axioms of congruence without explicitly referring to them.

• I'm just going to say this and then go back to reading the rest of this before I forget. The only reason I chose the wording I did was because it is straight out of the book we're using. It's Hilbert's euclidean parallel postulate. I'll continue reading now. Thanks for the post. – Fmonkey2001 Aug 5 '15 at 0:43
• @Fmonkey2001, Then quote exactly from Hilbert: "In a plane $\alpha$ there can be drawn through any point A, lying outside of a straight line $a$, one and only one straight line which does not inersect the line $a$. This straight line is called the parallel through the given point A." Important: the axiom part and the definition part are well separated. "The Foundations of Geometry", translated by E. J. Townsend) – zoli Aug 5 '15 at 0:52
• I apologize. I was using the definition that is straight out of our book. If it helps the book I'm using is Euclidean and Non-Euclidean Geometries by Greenberg. – Fmonkey2001 Aug 5 '15 at 0:59
• @Fmonkey2001, don't apologize, please. Perhaps I am a hair splitter (or what not). I am teaching myself :) But again: It is very important that neither the original Euclidean nor Hilbert's axiom use an a priory given concept of parallelism. From the point of view of proof writing it is a must that you distinguish between definitions implicitly given by axioms (point/ straight) and definitions based on already known concepts. ("don't intersect = parallel." – zoli Aug 5 '15 at 1:15
• ahhh, I understand now. That makes sense. I'm sure I wrote that down somewhere in my notes too. How quick I forget. I'll have to underline it and try to remember for next time. – Fmonkey2001 Aug 5 '15 at 1:17