Find the flaw in the attempted proof of the parallel postulate given by J. D. Gergonne The attempted proof: 
Given $P$ not on line $l$, line $PQ$ perpendicular to $l$ at $Q$, line $m$ perpendicular to $PQ$ at $P$ and point $A \neq P$ on $m$. Then, let $PB$ be the last ray between rays $PA$ and $PQ$ that intersects $l$, $B$ being the point of intersection. There exists a point $C$ on $l$ such that $Q * B * C$ (B is between $Q,C$). It follows that the ray $PB$ is not the last ray between rays $PA$ and $PQ$ that intersects $l$, and hence all rays between $PA$ and $PQ$ meet $l$. Thus $m$ is the only parallel to $l$ through $P$.
I can't seem to find it. I'm not sure if it is wrong because he is starting off with right angles, or something else. I feel that because he starts off with right angles, and this clearly only works for right angles, he is basically stating the parallel postulate for all the rays between the two lines as reasons for those not working. By stating and using the parallel postulate, he is being inconsistent. Am I on the right track?
 A: "$PB$ is not the last ray between rays $PA$ and $PQ$ that intersects $l$, and hence all rays between $PA$ and $PQ$ meet $l$"
The "hence" here is asserted without proof. In fact, it does not follow. All we have shown is that given any point $B$ there is a point $C$ beyond it.
For example, given any real number $b$ less than 2, there is another real number less than 2 but greater than $b$. Does it follow that all real numbers are less than 2?
Similarly, given any ray through $P$ intersecting $l$ at $B$, we have shown there is another ray through $P$ intersecting $l$ at $C$. Logically, it does not follow that all rays therefore intersect $l$.
There's not need for a counterexample; we just need to show the reasoning is flawed.
Like many of these attempted proofs that popped up throughout history, it's just a clever way to subtly hide the assumption of the parallel postulate by cloaking it in "geometric intuition". The "hence" is an appeal to the imagination; it is hard to imagine that such a ray between $PA$ and $PQ$ can exist without intersecting $l$, but that's not proof that it doesn't exist.
Edit: Ironically, people are downvoting my answer because they fell into the same trap Gergonne fell into. These would probably be the people (many of whom were otherwise skilled mathematicians) who accepted Gergonne's proof back in the day. What Gergonne proves by contradiction is that there is no last ray between $PA$ and $PQ$ that intersects $l$. That is not the same as saying there is no ray between $PA$ and $PQ$ that does not intersect l. They are not the same thing. With a little more structure you can think of it like this: he proves that the set of all the angles between $PQ$ and the rays that intersect $l$ has no greatest element. Logically it's still possible that it's bounded from above by some angle less than a right angle, so he hasn't proved what he set out to prove.
A: There is no "last ray PB"... even in Euclidean geometry.
The boundary between "rays from $P$ that intersects $\ell$" and "rays from $P$ that do not intersect $\ell$" turns out to be a ray that does not intersect $\ell$ affinely. However, it does meet $\ell$ "at infinity".
