In Proposition 20 of Book I of David E. Joyce's commentary on Euclid's Elements, we have the following fragment of a proof for the theorem that the shortest path between two points through a reflection is when the angle of incidence equals the angle of reflection:
We still have to show that the distance AE + EB is less than any distance AP + EP for any point P other than E that lies on the line CD. Let P be such a point and draw lines AP, BP, and B'P. Then by proposition I.20, above, AP + EP is less than AB'. But AB' = AE + EB', and EB' = EB, therefore AP + EP is less than AE + EB.
Please see link for the lead up to the proof.
This critical part of the proof confuses me in three different ways.
Why do we need to prove AE + EB is less than any distance AP + EP? Doesn't this require justification?
Which triangle does I.20 apply to? AP + EP don't seem directly related to AB' in a triangle where I.20 is applicable.
Then, why does this seem to conclude the opposite of what it set out to prove? "AE + EB is less than any distance AP + EP" vs. "therefore AP + EP is less than AE + EB"
Your confusion is well founded. It seems that there is a typo in the proof. It should say something along these lines:
"We still have to show that the distance $AE + EB$ is less than any distance $AP + PB$ for any point $P$ other than $E$ that lies on the line $CD$." Now we can use proposition I20 to show that $AP + PB=AP + PB'> AB'=AE + EB$