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Prove: I define the radius of three mutually externally tangent to be $d,e,f$ respectively. The circle with radius $x$ is internally tangent to all three circles. Then

$ddeeff+ddeexx+ddffxx+eeffxx = \\2(deffxx+ddeffx+deefxx+ddeefx+ddefxx+deeffx)$

[Reference: p.189-190 of The Changing Shape of Geometry.]

(Image in case you didn't see the google book here.)

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    @DayLateDon - Any proof would work for me, but more alternate way tend to be better.2012-05-02

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