$\angle POQ$ with angle bisector $OL$ is given. On its rays, $OP$ and $OQ$, points $A$ and $B$ are chosen, such that $OA < OB$. $a$ is the perpendicular line through $A$ to $OP$, and $a$ intersects $OL$ in $C$. $b$ is the perpendicular line through $B$ to $OQ$, and $b$ intersects $OL$ in $D$. $A_1$ is the image of $A$ to the line $OL$, $M$ is the midpoint of $CD$ and $N$ is the midpoint of $A_1B$. Show that $CA_1 \perp OQ$.
I think that $A_1$ lies on $OQ$ because $CA=C_1$ (is this enough to say this?). I think that the reflection of the line $OA$ is $OA_1$ and the reflection of $CA$ is $CA_1$, thus:
$OL = p$
$\sigma_p : O \to O$
$\sigma_p : A \to A_1$
$\sigma_p : OA \to OA_1$
$\sigma_p : C \to C$
$\sigma_p : CA \to CA_1$
$CA \perp OA \implies CA_1 \perp OA_1$
Is that okay?
The fact is :-
If we drop perpendiculars to the two arms (OP and OQ) of the angle (POQ) from any point (C, in this case) on the angle bisector, we get two congruent triangles (OCA and OCA'). CA' is automatically the reflected image of CA about OL.
Also, there is no need to use M and N.