See complete Problem 213 at:

gogeometry.com/problem/p213_triangle_inradius_common_tangents.htm

Level: High School, SAT Prep, College geometry

## Saturday, December 27, 2008

### Elearn Geometry Problem 213: Triangle, Incircle, Inradius, Semicircles, Common Tangents

Labels:
circle,
common tangent.diameter,
incircle,
tangency point,
triangle

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Let AF = AD = 2r1, FC = FE = 2r2 & BE = BD = 2r3

ReplyDeleteSemisum of sides, s = 2r1+2r2+2r3

Thus, r^2= (Area/s)^2 =(s-a)(s-b)(s-c)/s

where a=2r1+2r2,b=2r2+2r3 & c=2r3+2r1 which gives us, r^2 = (2r1*2r2*2r3)/(2r1+2r2+2r3)

= (4r1*r2*r3)/(r1+r2+r3)

or 1/r^2 = (r1+r2+r3)/(4r1r2r3) --------(1)

It's easy to see that, d^2=(r3+r1)^2-(r3-r1)^2 =4r3r1. Likewise, e^2=4r2r3 & f^2=4r1r2. Now, 1/d^2+1/e^2+1/f^2 = 1/4r1r2+1/4r2r2+3/4r3r1 = (r1+r2+r3)/(4r1r2r3)

By equation (1), 1/r^2 = 1/d^2 + 1/e^2 + 1/f^2

QED

Ajit: ajitaathle@gmail.com

By (1),1/r^2 = (r1+r2+r3)/(4r1*r2*r3) (r1+r2+r3)