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:

please follow the link to access it

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 :)


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 $BC.$

  • $\begingroup$ Oh okay. From what you have said, I added a point J on the bottom of the line BE and deduced that ∠FBJ = ∠FDB as BJ is a tangent (Alternate Segment Theorem). But how does this help with proving that CH and DF are parallel? I can't say yet that ∠FDC and ∠DCH are equal as I haven't proven that CH and DF are parallel. So what would I do now? $\endgroup$ – Cameron Choi Jul 18 '18 at 13:26
  • $\begingroup$ I was thinking about $\angle BFD$. $\endgroup$ – David K Jul 19 '18 at 1:02
  • $\begingroup$ Can you please elaborate on this as I still don't understand how you could prove that CH and DF are parallel using that angle? $\endgroup$ – Cameron Choi Jul 19 '18 at 8:20
  • 1
    $\begingroup$ The lines $CH$ and $DF$ both intersect the line $BF.$ If corresponding angles of the intersections are congruent, then the lines are parallel. $\endgroup$ – David K Jul 19 '18 at 11:24

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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.

  • $\begingroup$ One quick clarification: Q and P are both the centres of the circles, right? $\endgroup$ – Cameron Choi Jul 19 '18 at 2:22
  • $\begingroup$ @CameronChoi Yes, that is true. $\endgroup$ – Oldboy Jul 19 '18 at 4:29
  • $\begingroup$ @CameronChoi If you don't want to upwote the answer, please let me know what needs to be improved. $\endgroup$ – Oldboy Jul 19 '18 at 6:27
  • $\begingroup$ Sorry about taking a long time to upvote. Thank you for the alternative answer. $\endgroup$ – Cameron Choi Jul 19 '18 at 6:50

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