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The quadrilateral $ABCD$ is inscribed in a circle.The lines $AB$ and $CD$ meet at $E$,The diagonals meet at $F$.The circumcircle of $AFD$ and $BFC$ meet again at $H$.Prove that $\angle EHF$ is a right angle.

WLOG we may assume $AB\leq CD$. Suppose $AD$ intersects $BC$ at $G$. Notice that $G$ is the radical center of circles $ABC$, $AFD$ & $BFC$. So $F,G,H$ are collinear. Let $O$ be the center of circle $ABC$ and $R$ be its radius. Using power of point,
\[GO^2-R^2=GB\cdot GC=GF\cdot GH=GH^2-FH\cdot GH\quad (1)\]
$\angle GAC=180-\angle DAC=180-\angle DBC=\angle GHC$. So $G,A,H,C$ are concyclic.
\[FO^2-R^2=AF\cdot FC=GF\cdot FH=FH^2-FH\cdot GH\quad (2)\]
Subtracting (2) from (1), we get $GO^2-FO^2=GH^2-FH^2$. So by perpendicular lemma, it follows $GF\perp OH$. So done.

Comment: This problem is kinda obvious if anybody knows about Miquel point of a quad. Here $H$ is the Miquel point of quad $ACBD$ (in this order).

I will refer to other solutions that can be applied to solve the problem you need to solve.
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