Proposed Problem
See complete Problem 22 at:
gogeometry.com/problem/p022_right_triangle_incircles.htm
Level: High School, SAT Prep, College geometry
Sunday, April 5, 2009
Problem 22: Right triangle, Altitude, Perpendiculars, Inradii
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We've AD = c^2/b while AE = AD*cos(A)= AD*c/b = c^2/b * c/b = c^3/b^2 and DE = ccos(C)sin(C)= ac^2/b^2 where a,b,c have the usual meanings.
ReplyDeleteThus,r1=[c^3/b^2*ac^2/b^2]/[c^3/b^2+ ac^2/b^2 + c^2/b] or r1= ac^3/[(a+b+c)b^2]
Likewise, r2 = ca^3/[(a+b+c)b^2]
Therefore, r1+r2 =ac^3/[(a+b+c)b^2] +
ca^3/[(a+b+c)b^2] = ac(a^2+c^2)/[(a+b+c)b^2]
But a^2+c^2 = b^2 hence r1+r2 =acb^2/[(a+b+c)b^2]
=ac/(a+b+c) = r
Thus, r1+ r2 = r
Ajit: ajitathle@gmail.com
inradius for right tr
ReplyDeleter = ( a + b - c )/2
r1 = ( AE + ED - AD )/2
r2 = ( DF + FC - DC )/2
r1 + r2 = ( AE + ED - AD )/2 + ( DF + FC - DC )/2
r1 + r2 = ( AE + EB + BF + FC - ( AD + DC ))/2
r1 + r2 = ( AB + BC - AC )/2
r = ( AB + BC - AC )/2
r1 + r2 = r
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AED, ABC, DFC are similar.
ReplyDeleter1/r=AD/AC, r2/r = DC/AC
(r1+r2)/r=(AD+DC)/AC=1
r=r1+r2
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