Geometry problem : collinear points 
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
 A: 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.
A: 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 $ACBC'$.
