Proof Regarding Tangents There are two circles: $C1$ and $C2$. They touch internally at point B (note that C1 is inside of C2). C1 has a chord BC, which is produced to meet C2 at D. The tangent from C meets the common tangent from B at point E. There's a point F, which is where C2 and line CE intersect each other in a manner that causes E and F to be on opposite sides of BC. The line BF meets C1 at G, and CG meets the tangent from D at H. From this information, I need to show that CFDH is a parallelogram.
This is what I've done so far:

I know I still need to prove that CH and DF are also parallel to prove that CFDH is a parallelogram, but I'm a little stuck on how to do this. Also, I don't know if my diagram correctly represents the given information... Some help would really be appreciated.
Thanks :)
 A: The figure looks fine.
Step 1 should say $BE = CE.$
You're just one more application of the Alternate Segment Theorem away from finding the last angle you need.
Personally, I don't remember facts like the Alternate Segment Theorem (I had to look it up). I just remember a few formulas that relate the angle between two intersecting lines to the central angles of the arcs those two lines cut off on a circle. The Alternate Segment Theorem is just a consequence of the fact that each of the angles  $\angle EBC$ and $\angle BGC$ is half the central angle of the arc $BC.$
A: 
Check triangles $BQC$ and $BPD$. They are isosceles and they share one angle: $\angle QBC = \angle PBD$.
So these triangles are similar and their corresponding angles are equal. This means that $\angle BQC=\angle BPD$ which means that lines $QC$ and $PD$ are parallel. Tangets at points $C$ and $D$ are perpendicular to parallel lines $QC$ and $PD$ so these tangents are also parallel. Which means:
$$FC \parallel DH$$
In the same way you can show that lines $QG$ and $PF$ are parallel.
Angle $\angle CBG$ is inscribed angle for both circles so it means that their central angles must be equal: $\angle CQG = \angle DPF$. Triangles $CQG$ and $DPF$ are isosceles so they are similar and the corresponding angles must be equal. For example: $\angle CGQ=\angle DFP$. But lines $GQ$ and $PF$ are parallel and the second leg must also be paralel: $CG\parallel DF$ or:
$$CH \parallel FD$$
So opposite sides of $CFDH$ are parallel and this completes the proof.
The problem can also be solved more efficiently by using properties of homotethy, but I decided to provide an elementary proof instead.
