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Let $AB$ be the diameter of a circle $\omega$ and $H$ a point in $(BC)$ outside of the circle and let $l$ be the line from $ H$ perpendicular to $BA$ and $D$ a point on it Let $C = DA \cap \omega $, $G$ and $F$ be the tangency points from $D$ , and let $E$ be the pole of $BC$ and $H'$ the inverse of $H$ Show that $D$ , $H$ and $E'$ are collinear

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    Since $\triangle ABC \sim \triangle OEB \sim \triangle ADH$ and $|\overline{OH}||\overline{OH^\prime}| = |\overline{OA}|^2$, it's straightforward to show that $$\frac{|\overline{DH}|}{|\overline{HH^\prime}|} = \frac{|\overline{EB}|}{|\overline{BH^\prime}|}$$ so that $\triangle DHH^\prime \sim \triangle EBH^\prime$ and the result follows. – 2017-01-29

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Let $BG \cap CF = I$ and $BF \cap CG = K$. Then applying Pascal's Theorem on $GBACFGG$ we have that $D - I - H'$ are collinear. Again apply Pascal's Theorem on $FFCGGBF$ to obtain that $D - I - K$ are collinear. This means that $K$ lies on the line $D - I - H'$. For the last time apply Pascal's Theorem on $CCFBBGC$ to get that $E - I - K$ are collinear.

From all this we have that $D - K - E - I - H'$ are collinear, but more importantly $D, E, H'$ lie on the same line. Hence the proof.

UPDATE: To complete the proof I will prove that $H'$ lies on $GF$. We have that the pole of $H'$ is $DH$, therefore as $D$ lies on the pole of $H'$, we have that $H'$ lies on the pole of $D$, which is $GF$. Hence the proof.

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    This is a nice proof thanks ! actually, after proving $ D - I- H'$ are collinear we could apply brocard theorem on $BFCG$ – 2017-01-29
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It seems like one can also go without defining $F$ and $G$. You can extend $DB$ until it intersects the circle $\omega$ second time at point $C'$ and let line $AC'$ intersect $l = DH$ at $D'$. Then $D', \, B$ and $C$ are collinear (altitudes in triangle $ADD'$). Then one can show that $H'$ is the intersection of $CC'$ and $AB$ (again starting from the altitude argument). Then apply Pascal's theorem (degenerate version) to the degenerate hexagon $ACCC'BB$ concluding that the three intersection points $D, \, E$ and $H'$ are collinear.

The fact that $H'$ also lies on $FG$ follows from Brocard's theorem (I guess that's the name) which asserts that the segment $FG$ between the tangents $DF$ and $DG$ to circle $\omega$ passes through the intersection point $H'$ of the diagonals $AB$ and $CC'$ of the inscribed quad $ACBC'$.