Proving collinear points This problem is so hard that I cannot figure it out. I hope you guys can give me a small push on how to tackle this problem, as I have been thinking about this for, like a week. Here's the problem:
Problem
Given a circle with center O, radius R (denoted as (O; R)), and a point A exterior to (O). From A, draw 2 separate tangent line segments AB, and AC to (O) (B, C are tangent points). Let D be the intersection of the line segment AO, and (O). Choose a point E arbitrarily on the small arc BC. Let F, G be the intersection points of DE, with AB, and AC respectively. Let I be the intersection of CF, and BG; H be the intersection of AO, and BC.
Prove that H, I, and E are collinear points.
Picture

Any help would be greatly appreciated.

Why can't I embed pictures in my post? :((
 A: Let $X$ be the point where lines $BC$ and $FG$ meet. Let $J$ be the point where BC intersects $AE$. Let's forget about point $I$ for a moment. Here is the diagram:

Observe that $BD$ is the bisector of angle $HBA$, therefore $HB/BA=HD/DA$. Similarly, $HC/CA=HD/DA$. We see that points $B$, $C$ and $D$ all have the same ratio of distances from points $H$ and $A$. It follows that $(O)$ is an Apollonian circle of segment $HA$. Since $E$ is on this circle too, $HE/EA=HD/DA$. It follows that $ED$ is the bisector of angle $HEA$.
Now let us have a look at lines $EX$, $EB$, $EJ$, $EH$ and $EC$ and some of their cross-ratios. Let $f$ be the reflection across line $EX$. What we have established above is that $f(EJ)=EH$. It is also an easy exercise to see that $f(EB)=EC$ and $f(EX)=EX$. So, $f$ maps the four lines $EJ$, $EB$, $EC$ and $EX$ to lines $EH$, $EC$, $EB$ and $EX$ respectively. It follows that the corresponding cross-ratios are the same:
$$
(EJ,EB;EC,EX) = (EH,EC;EB,EX).
$$
Therefore, we have an equality for cross-ratios of points on line $BC$:
$$
(J,B;C,X) = (H,C;B,X).
$$
If we look at the perspective projection from line $BC$ to line $FG$ with center $A$, it sends points $J$, $B$, $C$ and $X$ to $E$, $F$, $G$ and $X$ respectively. Since perspective projections preserve cross-ratios, we have:
$$
(E,F;G,X) = (H, C; B, X).
$$
Now let us look at point $I$ (not shown on my diagram) where $BG$ and $FC$ intersect. Let $g$ be the perspective projection from line $BC$ to line $FG$ with center $I$. It is clear that $g(X)=X$, $g(B)=G$ and $g(C)=F$. $g$ preserves cross-ratios, therefore
$$
(H, C; B, X) = (g(H), F; G, X),
$$
and so
$$
(E,F;G,X) = (g(H), F; G, X).
$$
It follows that $g(H)=E$, which means that $H$, $E$ and $I$ are collinear, QED.
A: We do not need appolonian circles:
Let $P_{XY}$ be the point at infinity on $XY$. Let the parallel through $B$ to $EC$ meet $AC$ at $P$ and define $Q$ similarly. We have, 
$B(P_{EC}, G; E, C) = E(P, D; B, C) = E(Q, D; C, B) = C(P_{EB}, F; E, B)$
So if $BG \cap EC = X, CF \cap EB=Y$ then $XY \parallel BC \implies E, I, H$ are collinear.
Note: If one considers $EBC$, and takes the isogonals of the problem, it becomes similar to a problem used in the team selection tests for Bulgaria.
Furthermore, if the symmedian point of $BEC$ is $K$ then $K, I, B, C, A$ all lie on a conic. I believe this problem can thus be generalised and probably has already been.
