See complete Problem 200 at:

www.gogeometry.com/problem/p200_right_triangle_tangent.htm

Right Triangle, Incircle, Excircles, Points of Tangency, Inradius. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, November 3, 2008

### Elearn Geometry Problem 200: Right Triangle, Incircle, Excircles

Labels:
excircle,
incircle,
inradius,
right triangle,
tangency point,
tangent

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inradius in right triangle is r = (a + b - c )/2

ReplyDeletein this case

r = ( b + c - a )/2 => r = b/2 + c/2 - a/2 (a,hypothe)

r = b/2 + c/2 + a/2 - a/2 - a/2 ( add & subtract a/2 )

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

r = s - a

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name K small circle meet BC, L medium circle meet AC

T big circle meet BA

FC = LC ( tang from C )

FB + BK + KC = LA + r + EC

FB + BK = AH + r (KC = EC tg from C,LA = AH tg from A)

FB + BH + HD = AD + HD + r

FB + FB = r + r ( BH = FB tg from B, AD = r)

2 FB = 2r

FB = r

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BG = BT ( tg from B)

a + CG = AB + AT

a + CG = c + b - CM (AB = c, AT = AM = b - CM)

2CG = c + b - a

2CG = BD + r + EC + r - BD - EC (c=BD+r,b=EC+r,a=BD+EC

2CG = 2r

CG = r

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ADOE is square ( 4 angle = 90°, 2 side by side equal)

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