Let $h'$ the other-half part of the semicircle and let $M$ the midpoint of arc $AB$ in $h'$, now by homothety we can trivially see that $E,C,M$ and $F,D,M$ are colinear. Now by invertion at $(M, MA)$ we have $C \to E$ and $D \to F$ because $AB \to (h \cup h')$ therefore the cyclic follows from invert thus we are done .

By the Shooting Lemma lines $EC$ and $FD$ pass through the midpoint $M$ of arc $AB$. The rest is just an angle chase $$\angle MCD=\frac{1}{2} \overarc{EA}+\frac{1}{2}\overarc{MB}=\frac{1}{2} \overarc{EA}+\frac{1}{2}\overarc{AM}=\frac{1}{2}\overarc{EM}=\angle EFM$$

Attachments:

Well in fact, part of the Shooting Lemma tells you $MC \cdot ME=MA^2=MB^2$ so that you get $MC \cdot ME=MD \cdot MF$ and are done immediately.

What is the Shooting Lemma?

Kiewi wrote: What is the Shooting Lemma? Pretty much exactly the first sentence of #3, i.e. in the situation of the two touching circles with a tangent line, the line through the two tangency points passes through the midpoint of the arc. Often, the fact in #4 is also considered part of the lemma.

Invert about the midpoint of the other semicircle in the circle which $h$ is a part of, the result follows immediately.