Proof involving the rotation of a line through angle $\alpha$ Show that if a line l is rotated about any point O through an angle with measure $\alpha$ to a new line l$’$, then lines l and l$’$ intersect in an angle with measure $\alpha$.
I have been attempting to solve this problem, but am not too sure how to go about it. So far I have done a quick sketch and used some basic angle properties, but have not got very far.

Any hints would be greatly appreciated.
side note: if possible can someone explain how to format the phrase l prime in math Jax. Thanks.
 A: Let $\{A,B\}\subset l$ and $R_O^{\alpha}A=A',$ $R_O^{\alpha}B=B'$ and $A'B'\cap AB=\{K\}.$
Let $K$ be placed between $A$ and $B$.
Thus, since $\Delta OAB\cong\Delta OA'B',$ we obtain:
$$\measuredangle B'KB=180^{\circ}-\measuredangle A'KA=180^{\circ}-(360^{\circ}-\alpha-\measuredangle OAK-\measuredangle OA'K)=$$
$$=-180^{\circ}+\alpha+\measuredangle OB'A'+B'OA'+180^{\circ}-\measuredangle OBA-\measuredangle BOA=\alpha.$$
A: Let C be the intersection between l and l'. We take any points A and B from line l, such that C is between A and B. On the other hand, A' and B' are the images of A and B, respectively. By properties of rotation, AB=A'B', OA=OA', and OB=OB'. Hence, by congruency SSS $\triangle$OAB$\cong$$\triangle$OA'B', as shown below. Besides, there is an intersection between segments OA and A'B', which we call point D.

By congruence, $\angle$OAB=$\angle$OA'B', and also we have that $\angle$ADC=$\angle$A'DO. Thus, by similarity AA, $\triangle$ADC$\sim$$\triangle$A'DO. Then, $\angle$ACA'=$\angle$AOA'.
