Here is a proposition that I saw presented in a plane geometry textbook. The argument in the textbook was unusually awkward. I think that there is a simple argument for it. I would appreciate comments to the demonstration that I am providing.
$\overline{AB}$ is a chord in a circle, and $\ell$ is a line that is tangent to the circle at $P$. $\ell$ is parallel to $\overline{AB}$ if, and only if, $\arc{AP} = \arc{BP}$.
Demonstration of "Only If"
$\overline{AB}$ and $\ell$ are parallel, and $k$ is the line through $P$ that is perpendicular to $\ell$. $k$ contains the center $O$ of the circle, and so, $k = \dvec{OP}$. $k$ is perpendicular to $\dvec{AB}$, too. If $M$ is the midpoint of $\overline{AB}$, $\dvec{OM}$ is perpendicular to $\overline{AB}$. There is only one line through $O$ that is perpendicular to $\overline{AB}$. So, $k = \dvec{OM} = \dvec{OP}$.
$\triangle{AOM} \cong \triangle{BOM}$. If $\overline{PQ}$ is a diameter of the circle, \begin{equation*} {\mathrm{m}}\angle{AOQ} = {\mathrm{m}}\angle{AOM} = {\mathrm{m}}\angle{BOM} = {\mathrm{m}}\angle{BOQ} . \end{equation*} By the definition of angular measure, ${\mathrm{m}}\bigl(\arc{AQ}\bigr) = {\mathrm{m}}\bigl(\arc{BQ}\bigr)$. So, \begin{equation*} {\mathrm{m}}\bigl(\arc{AP}\bigr) = 180 - {\mathrm{m}}\bigl(\arc{AQ}\bigr) = 180 - {\mathrm{m}}\bigl(\arc{BQ}\bigr) = {\mathrm{m}}\bigl(\arc{BP}\bigr) . \ \rule{1.5ex}{1.5ex} \end{equation*}
The following code can be compiled by TikZ to render a diagram associated with this argument.
\noindent \hspace*{\fill} \begin{tikzpicture}
\coordinate (O) at (0,0); \draw[fill] (O) circle (1.5pt); \draw[name path=circle] (O) circle (2);
%Points A and B are drawn on the circle. Radii $\overline{OA}$ and $\overline{OB}$ are drawn. \coordinate (A) at (180:2); \node at ($(A)!-3mm!(O)$){$A$}; \coordinate (B) at (70:2); \node at ($(B)!-3mm!(O)$){$B$}; \draw (A) -- (B); \draw (O) -- (A); \draw (O) -- (B);
%The diameter of the circle is drawn through the midpoint M of $\overline{AB}$. %The endpoints are labeled P and Q. \coordinate (M) at ($(A)!0.5!(B)$);
%The label for M is typeset. \coordinate (label_for_M_above) at ($(M)!1cm!(A)$); \coordinate (label_for_M_right) at ($(M)!1cm!(B)$); \path (label_for_M_above) -- (label_for_M_right); \coordinate (midpoint_on_line_segment_to_position_M) at ($(label_for_M_above)!0.5!(label_for_M_right)$); \node at ($(M)!4mm!(midpoint_on_line_segment_to_position_M)$){$M$};
%The diameter $\overline{PQ}$ is drawn. \path[name path=path_to_locate_P] (M) -- ($(M)!3.5cm!(O)$); \coordinate[name intersections={of=path_to_locate_P and circle, by=P}]; \node at ($(P)!-3mm!(M)$){$P$}; \draw[fill] (P) circle (1.5pt);
\path[name path=path_to_locate_Q] (M) -- ($(M)!-1cm!(O)$); \coordinate[name intersections={of=path_to_locate_Q and circle, by=Q}]; \node at ($(Q)!-3mm!(M)$){$Q$};
\draw (P) -- (Q);
%A right-angle mark is drawn at M. \coordinate (U) at ($(M)!4mm!-45:(O)$); \draw[dashed] (U) -- ($(M)!(U)!(O)$); \draw[dashed] (U) -- ($(M)!(U)!(A)$);
%The label for O is typeset. \coordinate (label_for_O_above) at ($(O)!1cm!(B)$); \coordinate (label_for_O_right) at ($(O)!1cm!(P)$); \path (label_for_O_above) -- (label_for_O_right); \coordinate (midpoint_on_line_segment_to_position_O) at ($(label_for_O_above)!0.5!(label_for_O_right)$); \node at ($(O)!3mm!(midpoint_on_line_segment_to_position_O)$){$O$};
%Line segments $\overline{OA}$ and $\overline{OB}$ are marked with "|". \draw ($($(O)!0.5!(A)$)!3pt!90:(O)$) -- ($($(O)!0.5!(A)$)!3pt!-90:(O)$); \draw ($($(O)!0.5!(B)$)!3pt!90:(O)$) -- ($($(O)!0.5!(B)$)!3pt!-90:(O)$);
%Line segments $\overline{AM}$ and $\overline{BM}$ are marked with "||". \draw ($($($(A)!0.5!(M)$)!1pt!(A)$)!3pt!90:(A)$) -- ($($($(A)!0.5!(M)$)!1pt!(A)$)!3pt!-90:(A)$); \draw ($($($(A)!0.5!(M)$)!1pt!(M)$)!3pt!90:(A)$) -- ($($($(A)!0.5!(M)$)!1pt!(M)$)!3pt!-90:(A)$); \draw ($($($(B)!0.5!(M)$)!1pt!(B)$)!3pt!90:(B)$) -- ($($($(B)!0.5!(M)$)!1pt!(B)$)!3pt!-90:(B)$); \draw ($($($(B)!0.5!(M)$)!1pt!(M)$)!3pt!90:(B)$) -- ($($($(B)!0.5!(M)$)!1pt!(M)$)!3pt!-90:(B)$);
%The tangent line $\ell$ to the circle at P is drawn. \draw[-latex] (P) -- ($(P)!3.5cm!90:(O)$); \draw[-latex] (P) -- ($(P)!3.5cm!-90:(O)$); \node at ($(P)!{3.5cm+0.3cm}!-90:(O)$){$\ell$};
%The angle mark for $\angles{AOQ}$ is drawn. It is marked with "|". \draw[draw=blue] let \p1=($(O)-(Q)$), \n1={atan(\y1/\x1)} in ($(O)!0.4cm!(A)$) arc (180:{\n1+180}:0.4); \draw[blue] let \p1=($(O)-(Q)$), \n1={atan(\y1/\x1)} in ($(O) +({0.5*(180+(\n1+180))}:{0.4cm-3pt})$) -- ($(O) +({0.5*(180+(\n1+180))}:{0.4cm+3pt})$);
%The angle mark for $\angles{BOQ}$ is drawn. It is marked with "|". \draw[draw=blue] let \p1=($(O)-(Q)$), \n1={atan(\y1/\x1)} in ($(O)!0.4cm!(Q)$) arc ({\n1+180}:70:0.4); \draw[blue] let \p1=($(O)-(Q)$), \n1={atan(\y1/\x1)} in ($(O) +({0.5*((\n1+180)+70)}:{0.4cm-3pt})$) -- ($(O) +({0.5*((\n1+180)+70)}:{0.4cm+3pt})$);
\end{tikzpicture} \hspace{\fill}
I think this version of the approach might be a little bit simpler, although it is essentially the same idea.
By the inscribed angle theorem, the claim is equivalent to showing that $\angle ABP = \angle BAP$. This is in turn equivalent to showing that the perpendicular to $\mathit{l}$ through $P$ bisects $AB$.
So, take your line $k$ perpendicular to $\mathit{l}$ at $P$, and note (like you did) that $k$ goes through the center of the circle $O$ and is also perpendicular to $AB$. Now look at the triangle $OAB$. Since it is isosceles with base $AB$ that means that $k$ must bisect $AB$.