Proof of Monge's circle theorem

[wakko101]

Homework Statement

Consider 2 circles. For every couple of circles consider their two common external tangent lines and take their point of intersection. Prove that these 3 points of intersection belong to one line.

Homework Equations

Menelaus' theorem, possible Ceva's theorem as well.

The Attempt at a Solution

I've seen a few proofs that extend the circles to spheres, but we're dealing with simply the plane. I'm just not sure how to begin here. I'm wondering if there is something about the geometry of the circles and their relationship that I need to know to get started. Any assistance would be greatly appreciated.

Cheers,
W. =)

 

[mighty2000]

Dear W.,

Thank you for bringing up this interesting problem. Let's start by considering the two circles as shown in the diagram below:

[image of two circles with external tangent lines]

We can label the points of intersection of the circles as A, B, C, and D. Now, let's draw the two common external tangent lines, as shown in the diagram. We can label the points of intersection of these lines with the circles as E, F, G, and H.

Next, let's draw a line connecting points A and G, and another line connecting points B and H. These two lines will intersect at a point, which we can label as X. Similarly, we can draw a line connecting points C and F, and another line connecting points D and E. These two lines will also intersect at a point, which we can label as Y.

Now, let's consider the triangle ACG. By Menelaus' theorem, we have:

AC/CG * GX/XF * FY/AY = 1

Similarly, for the triangle BHD, we have:

BD/DH * HY/YG * GX/XB = 1

Since the two tangent lines are also common external tangents, we know that AC = BD and CG = DH. Therefore, we can substitute these values into the equations above, giving us:

BD/CG * GX/XF * FY/AY = 1

AC/DH * HY/YG * GX/XB = 1

Now, we can rearrange these equations to get:

GX/XF = CG/AY * FY/BD

GX/XB = DH/YG * HY/AC

Since we know that CG = DH and AY = FY, we can simplify these equations to get:

GX/XF = FY/BD

GX/XB = HY/AC

Since XF and XB are both tangent lines to the circles, we know that FY = HY and BD = AC. Therefore, we can rewrite the equations as:

GX/XF = HY/AC

GX/XB = HY/AC

Since both equations are equal to HY/AC, we can set them equal to each other:

GX/XF = GX/XB

Therefore, we can conclude that XF and XB intersect at the same point, which is X. Similarly, we can show that Y is the point of intersection for YF and YE. Therefore, we have shown that all three points of intersection