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

Labels:
altitude,
inradius,
perpendicular,
right triangle

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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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