# The Eyeball Theorem

Given two circles (A) and (B). Let the tangents from A to (B) meet (A) in points
P and Q, and the tangents from B to (A) meet (B) in points R and S. Then the
eyeball theorem says that PQ = RS. Or equivalently that PQSR is a rectangle. Some proofs are given on
Alexander Bogomolny's Cut the
Knot site. Here is my proof.

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#### Proof

First note that AMB, ANB, ALB and AKB are right triangles
with AB as hypothenuse, so that the vertices all lie on a circle with diameter AB.
Now extend PR to TU with T on (A) and U on (B). Let TM and UN meet in V.

Note that angle(MAN) = angle(MBN) and thus angle(VTU)
= angle(MTP) = angle(MAN)/2 = angle(MBN)/2 = angle(RUN) = angle(VUT), so
triangle TUV is isosceles with VT=VU.

We also see that angle(MBN) = angle(VTU) + angle(TUV)
so that angle(MBN) and angle(NVM) add to 180 degrees, so that V lies on the
circle with diameter AB.

Now note that angle(TVA)=angle(KVA), as these angles
intercept congruent chords, and angle(UVB) = angle(LVB). By reflection
through AV and BV respectively, we see that in fact VT=VK=VL=VU.

Let W be the reflection of V through AB. Triangle WMN
is isosceles just as triangle KLV. Hence angle(WVN) = angle(WMN) = angle(WNM)
= angle(WVM), so VW bisects angle(TVU). So VW is perpendicular to PR as
angle bisector of the apex of an isosceles triangle. And thus PR is parallel
to AB. Similarly QS is parallel to AB. And PQSR is a rectangle.

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