Eight Equal Tangents

Tangents from a point to a circle are equal. Tangents to two circles from a point on their radical axis are also equal. The four common tangents to two circles (when all four are available) supply a slew of equal segments. An applet below provides an illustration for a less known fact: the emphasized segments - all eight of them - are indeed equal.

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Solution

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Copyright © 1996-2017 Alexander Bogomolny

Let show, for example, that in the diagram below, EK = FL:

AB= AK + BK
 = EK + FK
 = EK + (EK + EF)
 = 2·EK + EF.

Similarly,

CD= CL + DL
 = EL + FL
 = (FL + EF) + FL
 = 2·FL + EF.

Since AB = CD we also have

2·EK + EF = 2·FL + EF

and, finally, EK = FL.

We could similarly show that in the diagram below also EK = FL.

However, the derivation above also works in this case provided some of the segments are allowed to be negative.

References

  1. R. Honsberger, More Mathematical Morsels, MAA, 1991, pp. 4-5.

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Copyright © 1996-2017 Alexander Bogomolny

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