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?
Best Answer
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
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
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.