From a point $M$ outside a circle $(O)$, draw two tangents $MA$ and $MB$ and a line $MCD$ cuts through the circle $(O)$ (positions of points $C$, $D$ are in the figure below). Let $H$ be the intersection of $MO$ and $AB$. Prove that $\widehat{ADH}=\widehat{CDB}$.
Attempt:
Let $MO$ intersects the circle $(O)$ at $E$ and $F$ (positions are in the figure above).
Using the Thales' theorem, we can prove that $\widehat{ECF}=\widehat{EHB}=90^\circ$, then $\widehat{ECI}+\widehat{EHI}=180^\circ \Rightarrow$ $ECIH$ is a cyclic quadrilateral.
We can prove that $\Delta{MCH}$ and $\Delta{MOD}$ are similar, so $\widehat{OHD}=\widehat{MHC} \Rightarrow 90^\circ-\widehat{OHD}=90^\circ-\widehat{MHC} \Rightarrow \widehat{BHD}=\widehat{BHC}$
Because $ECIH$, $AECD$, $ACBD$ are cyclic, $\widehat{BHD}=\widehat{BHC}=\widehat{IEC}=\widehat{CAD} \Rightarrow 180^\circ-\widehat{BHD}=180^\circ-\widehat{CAD} \Rightarrow \widehat{DHA}=\widehat{DBC}$.
We have $\widehat{DHA}=\widehat{DBC}$ and $\widehat{DCB}=\widehat{DAH}$ because of inscribed angle, so $\Delta{DHA}$ and $\Delta{DBC}$ are similar, so $\widehat{ADH}=\widehat{CDB}$.
My question is: I think the only flaw in this proof is that I have to prove that $DE,CF,AB$ all intersect at one point called $I$. Is this true? Note that I actually know how to prove these three lines concurrent, I just need verification.