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
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.
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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