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### Two Circles, Four Tangents, Collinear Midpoints

```
Date: 12/20/98 at 05:19:04
From: hai trung ho
Subject: Geometry

Given two circles that do not touch, there are four distinct tangents
common to both circles. Prove that the midpoints of the tangents are
collinear.
```

```
Date: 12/24/98 at 07:25:56
From: Doctor Floor
Subject: Re: Geometry

Hello Hai Trung Ho,

To make the explanation easier I have made a picture for you:

We will repeatedly use the fact that two tangents to one circle from
the same point have the same length, and of course we will use the
symmetry in the line DA.

As a first consequence we can introduce the following names for
segment lengths:

p: = OT = TI = EN = HN
q: = BS = SQ = LR = RK

We will show first that in fact two names were not necessary, because
p = q.

To see this, realize that TL = TQ = OQ-p. And thus IL = TL-p = OQ-2p.

But also RI = RH = KH-q = OQ-q. And thus IL = RI-q = OQ-2q.

Since IL = OQ-2q = OQ-2p, apparently p = q.

But then the midpoint of HK is equal to the midpoint of NR, because
from both the left and the righthand side the same length is taken. In
the same way, the midpoint of EB is equal to the midpoint of NS.

Since the two segments NR and NS both start from point N, the line
connecting their midpoints must be parallel to RS and this line must be
perpendicular to DA, since RS is.

Conclusion: the line through the midpoints of HK and EB is
perpendicular to DA.

Because of the symmetry of the complete figure in DA, this conclusion
implies that the line through the midpoints of IL and OQ is
perpendicular to DA too, having the same point of intersection with
DA. But then the four midpoints must be on one and the same line! And

Best regards,

- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles
High School Euclidean/Plane Geometry
High School Geometry

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