I am currently learning proofs related to geometry, which is why I have been looking at this problem (I hope it is okay to link to external websites). Here is the problem: Given a point $A$ outside of a circle. Two tangents go through the point $A$ and touch the circle at $B$ and $C$ respectively. The center of the line $BC$ is $M$. Given a secant through $A$ which crosses the circle at $D$ and crosses $BC$ at $E$, show that $\angle BDM=\angle EDC$.
Users have posted solutions but they seem rather incomplete to me because I can't follow those solutions posted by other users.
So I wanted to show my idea/ approach:
It suffices to show that the arc $BG$ and $CF$ are equal in length. That can be shown by showing that $\angle AMD=\angle HMF$. So $AH$ has to be an external angle bisector of $\angle DMF$. Of course $AH$ is the (internal) angle bisector of $\angle BAC$ because $M$ lies in the middle of $BC$. Unfortunaletly that's were I can't make any meaningful progress because I don't know to show how that $\angle AMD=\angle HMF$ is true.
Tips and help would be appreciated!
$AO$ is the perpendicular bisector of $BC$.
$\triangle ABM\sim \triangle AOB$ and thereafter $AM\cdot AO=AB^{2}$. Also since $AD\cdot AL=AB^{2}$, $AD\cdot AL=AM\cdot AO$ and thereafter quadrilateral $LOMD$ is cyclic.
Now by simple angle chasing it will follow that $\angle DMC=\angle DBL$ and thereafter $\triangle BDL\sim \triangle MDC$ and hence $\angle BDL=\angle MDC$.