Proof that Corresponding Angle Postulate $\iff$ Playfair's Postulate

Question

I will state both postulates here:

Corresponding Angle Postulate: If two lines are cut by a transversal, then the lines are parallel if and only if the corresponding angles are congruent.

And

Playfair's Postulate: Given a line and a point not on that line, there is exactly one line passing through the point that is parallel to the given line.

I have looked online for a proof that they are equivalent but I am struggling to find one. We would have to prove and if an only if statement. Corr$\angle$ implies Playfair and Playfair implies Corr$\angle$.

Any guidance on what a proof of these two would look like?

Answer 1

Frist, let's show that Equal Corresponding Angles implies Parallel Lines

Let $\overline{AB}$ and $\overline{CD}$ be infinite straight lines. Let $\overline{EF}$ be a transversal that cuts them. Let at least one pair of corresponding angles be equal. WLOG, let $\angle EGB = \angle GHD$. By the Vertical Angle Theorem: $$\angle GHD = \angle EGB = \angle AGH$$ Thus by Equal Alternate Angles implies Parallel Lines: $$\overline{AB} \| \overline{CD}$$

Now we can do our iff proof:

Euclid's fifth postulate implies Playfair's axiom

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Given a line $\ell$ and a point $P$ not on $\ell$, construct a line $t$, perpendicular to the given one through $P$, and then a perpendicular to this perpendicular at the point $P$. This line is parallel because it cannot meet $\ell$ and form a triangle, which is stated in Book 1 Prop 27. Now it's clear that no other parallels exist. If $n$ was a second line through $P$, then $n$ makes an acute angle with $t$ (since it is not the perpendicular) and the hypothesis of the fifth postulate holds, and so $n$ meets $\ell$.

Playfair's axiom implies Euclid's fifth postulate

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We will show the contrapositive. So assume Euclid's fifth postulate is false, that is, there is a line $t$ intersecting lines $l,l'$ such that the sum of the interior angles on one of the sides of $t$ is less than two right angles, yet $l,l'$ do not meet on that side of $t$. They cannot meet on the other side either, since on that other side the sum of interior angles is greater than two right angles, and even without Euclid's fifth postulate the sum of two angles in a triangle cannot exceed two right angles. Therefore $l,l'$ are parallel.

We may also construct another line $l''$ through the point $P$ on $l'$ where the transversal $t$ meets it, such that for this line the sum of its internal angles on the same side of $t$ is equal to two right angles, and this line $l''$ is parallel to $l$ (also without Euclid's fifth postulate). So we now have two distinct parallels to $l$ through $P$ and have arrived at the negation of the Playfair axiom. ‍

Answer 2

First of all both postulates must be restated in a weaker (but equivalent) form because by Euclid I-27 we know that lines forming two congruent alternate (or corresponding) angles with a transversal are parallel between them. [Note added after some comments: This theorem DOES NOT REQUIRE THE PARALLEL POSTULATE.] Hence the existence of a parallel line through a given point is guaranteed by that theorem and needn't be postulate. We are only concerned with the uniqueness of such line.

Corresponding Angle Postulate: If two parallel lines are cut by a transversal, then the corresponding angles are congruent.

Playfair's Postulate: Given a line and a point not on that line, there is only one line passing through the point that is parallel to the given line.

• Let's prove by contradiction that Playfair's Postulate implies Corresponding Angle Postulate.

Given a line $r$ and a point $P\notin r$, suppose there exists a line $a$ through $P$ parallel to $r$ forming with a transversal $t$ different corresponding angles. We can then construct another line $b\ne a$ through $P$ forming with $t$ congruent corresponding angles. By Euclid I-27 it follows that $b\parallel r$. But this violates Playfair's Postulate because there would be two distinct lines through $P$ parallel to the same line $r$. $\square$

• Let's prove by contradiction that Corresponding Angle Postulate implies Playfair's Postulate.

Given a line $r$ and a point $P\notin r$, suppose there exist two lines $a$, $b$ through $P$, both parallel to $r$. Construct any transversal through $P$ intersecting $r$. As $a\ne b$ at least one of them will form different corresponding angles with $r$, thus violating the Corresponding Angle Postulate. $\square$

Answer 3

This is a proof found on Euclidean and Non-Euclidean Geometries/Development and History, by Marvin Jane Greenberg W. H. Freeman; 4th edition (September 28, 2007), page 174.

It makes use of axioms and theorems that are part of Hilbert's "rework"-if the term is applicable-of Euclidean geometry. I will only provide their statements. For a more comprehensive treatment of the subject I strongly suggest reading the book.

Since detaching a proof from a book can be like cutting part of a painting-we lose the context and the "flow" of the text-I tried to adapt it a bit to make it smoother, reconstructed the figure and changed some symbols, but the arguments are all here.

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There is also a good explanatory section on the Playfair's postulate article of Wikipedia about this equivalence.

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We will use Reductio ad Absurdum. First, we assume the Playfair's postulate holds ($\Leftarrow$). The situation of the Corresponding Angle Postulate (Euclid V), is shown in the Figure below.

By hypothesis we have: $\hat x+\hat y\lt 180^{\text o}$ and $\hat x+\hat z= 180^{\text o}$ (supplementary angles).

Hence, $\hat y\lt 180^{\text o}-\hat x=\hat z$.

There is a unique ray $\vec {B'C'}$ such that $\hat z$ and $\hat {C'B'B}$ are congruent interior angles due to axiom C-4.

By Theorem $4.1$, $\overleftrightarrow {B'C'}$ is parallel to $l$. Since $m\neq\overleftrightarrow {B'C'},m$ meets $l$ by Playfair's postulate.

To conclude, we must prove that $m$ meets $l$ on the same side of $t$ as $C'$.

Assume, on the contrary, that they meet at a point $A$ on the opposite side.

Then $\bar y$ is an exterior angle of $\triangle ABB'$. Yet it is smaller than the remote interior angle $\hat z$. This contradiction of Theorem 4.2 proves the Corresponding Angle Postulate (Euclid V).

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Now for the opposite $(\Rightarrow)$ direction.

Conversely, we assume that the Corresponding Angle Postulate (Euclid V) holds, and refer to the Figure below for Playfair's postulate.

Let $t$ be the perpendicular to $l$, through $P$, and $m$ the perpendicular to t through P.

We know that $m||l$ from the Corollary to 4.1.

Let $n$ be any other line through $P$. We must show that $n$ meets $l$.

Let $\hat x$ be the acute angle $n$ makes with $t$ (which angle exists because $n\neq m$).

Then we have $\hat x +\hat{PQR}\lt 90^{\text o}+90^{\text o}=180^{\text o}.$

Thus the hypothesis of the Corresponding Angle Postulate (Euclid V) is satisfied.

Hence $n$ meets $l$, proving Playfair's postulate.

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We made use of these:

Theorem 4.1

In any Hilbert plane, if two lines cut by a transversal have a pair of congruent alternate interior angles with respect to that transversal, then the two lines are parallel.

Corollary to 4.1

Two lines perpendicular to the same line are parallel.

Theorem 4.2.

In any Hilbert plane, an exterior angle of a triangle is greater than either remote interior angle.