Similar Triangles and the Power of a Point

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Let a line $k$ be tangent to a circle $o$ at a point $D$. Let $A$ and $B$ be points on $o$ such that $AB$ is parallel to $k$. Let $C$ be a point on $k$ such that the line segments $CA, CB$ intersect $o$ in points $E$ and $F$. Prove that the line $EF$ intersects $CD$ in a point $M$ that is the centre of $CD$. Hint: prove that $\angle ECD=\angle MFC$ and find similar triangles to compute $|M C|$. Then use the power of $M$ with respect to $o$ to compute $|M D|$

I tried using the hint but I could not prove the angle part. I tried using Thales with respect to the point $F$ and the segment $MC$ and line $AB$ but I did not get anything,

Any help would be appreciated,

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Thanks

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$ABFE$ is a cyclic quadrilateral, then (by property of cyclic quadrilateral) $$\angle EAB =\angle EFC$$ Also $\angle EAB =\angle MCE$, because $AB$ is parallel to $DC$. From this we have that $MCE \sim MFC$. Hence $$\frac{MC}{MF}=\frac{ME}{MC}$$ This implies that $ME\cdot MF=MC^2$. From here is easy conclude the problem.