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One excircle to another (Posted on 2021-03-12) Difficulty: 3 of 5
It turned out for some triangle with sides a, b and c, that a circle of radius r = (a+b+c)/2 touches side c and extensions of sides a and b. Prove that a circle of radius (a-b+c)/2 is tangent to a and the extensions of b and c.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution Comment 1 of 1
Letting s be the semiperimeter s=(a+b+c)/2, which is the given value of rc.

The formula for the radius of the excircle to side c is

rc = sqrt(s(s-a)(s-b)/(s-c))

Substituting the given value.

s = sqrt(s(s-a)(s-b)/(s-c))

sqrt(s(s-c))=sqrt((s-a)(s-b))

sqrt(s(s-b)(s-c)/(s-a))=s-b

This is the radius of excircle to side a.

=(a+b+c)/2 - b

=(a-b+c)/2

As required.

  Posted by Jer on 2021-03-13 08:09:00
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