See complete Problem 207 at:

gogeometry.com/problem/p207_right_triangle_hypotenuse.htm

Right Triangle, Hypotenuse, Inradius, Exradius relative to the hypotenuse. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Saturday, November 15, 2008

### Elearn Geometry Problem 207: Right Triangle, Inradius, Exradius, Hypotenuse

Labels:
exradius,
hypotenuse,
inradius,
right triangle,
tangency point

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let s be the semi perimeter and r be the in radius and ra be the exradii of excircle relative to the side BC and let a,b,c be lenghths of sides BC,CA,AB we have some known relations AE=AD=s-a and AF=AG=s so EF=AF-AE=a and DG=AG-AD=a hence EF=DG=BC=a. we know that ra=Ar of TrABC/(s-a) and r=Ar of TrABC/s . so ra-r=Ar of TrABC.a/s(s-a)= Tan A/2 .a = Tan 45.a = a (Angle A = 90) HENCE ra-r=a1=a2=a. that means BC=EF=DG=ra-r

ReplyDeleteSee the

ReplyDeletedrawing(1) AF=ra=r+EB+BF=r+a2

(2) AG=ra=r+DC+CG=r+a1

(1-2) a1=a2

(1+2) ra=r+a1=r+a2 => a1=a2=ra-r

Define X as the intersection of BC and circle ra

Define Y as the intersection of BC and circle r

CD and CY tangents to circle r => CD=CY

BE and BY tangents to circle r => BE=BY

a=CY+BY=> (3) a=CD+BE

CG and CX tangents to circle ra => CG=CX

BF and BX tangents to circle ra => BF=BX

a=BX+CX=> (4) a=BF+CG

(3+4) 2a= CD+BE+ BF+CG=CD+CG+BE+ BF=a1+a2

=> a=a1=a2

Therefore a=a1=a2=ra-r