Euclidean Geometry, Question involving reflection, Homothety.

Question

So, let's say there is a line $l$ (here it is $XY$) , $A$ and $B$ are random points on one side of $l$, and $A'$ and $B'$ are their reflections across $l$, I need to prove $AB'$, $A'B$ concur at $l$

Now, I was reading a solution to a problem which I couldn't solve when the author said that this is obvious, as I was low on confidence at that time, I couldn't see it at all, even though this seems completely trivial, so I tried to prove it

I have just learned Homothety, so this was my proof;

let $AB'$ intersect $A'B$ at $O$

Taking a homthety at $O$ which sends $A$ to $B'$

It will send $A'$ to $B$

And hence, it must send $P_1$ to $P_2$

Hence, $P1-O-P2$

Now, I have a couple questions,

**Is this proof correct?

Can you please tell me a normal proof (without homothety)**

Sorry for such a trivial question, idk why I am not able to think :(

Answer 1

This is a proof without homothety.

Let $AB'$ intersect $l$ at point $P$.

$l$ is the perpendicular bisector of $AA'$ and $BB'$.

$\angle APX=\angle B'PY$

Hence, $\angle A'PX=\angle APX=\angle B'PY=\angle BPY$ and thereafter points $A'$, $P$ and $B$ are collinear.