# In center-excenter configuration in a right angled triangle

My question is:

Given triangle ABC , where angle C=90 degrees. Prove that the set { s , s-a , s-b , s-c } is identical to { r , r1 , r2 , r3 }.

*s=semiperimeter , r1,r2,r3 are the ex-radii.

Any help to solve this would be greatly appreciated.

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and presumably $r$ is the inradius, which should be equal to $s-c$. – Henry Jun 10 '12 at 11:32
@Henry how did you get that? – mgh Jun 11 '12 at 9:57
I drew an example and measured the various lengths. $s$ is the ex-radius opposite $C$. It confirmed your statement, though it is not a proof. – Henry Jun 11 '12 at 16:04
The result is (the converse of) Amy Bell's proposition 3. What she calls "the formulas for the exradii" can be found on Wikipedia – Henry Jun 11 '12 at 21:38
@Henry thanks.. – mgh Jun 11 '12 at 22:13

This problem have some interesting fact behind so I use pure geometry method to solve it to show these facts:

first we draw a picture.

$\triangle ABC$, $I$ is incenter, $A0,B0,C0$ are the tangent points of incircle. $R1,R2,R3$ is ex-circle center,their tangent points are $A1,B1,C1,A2,B2,C2,A3,B3,C3$.

FOR in-circle, we have $CA0=CB0=s-c,BA0=BC0=s-b,AB0=AC0=s-a$,

now look at circle $R2, BC2$ and $BA1$ are the tangent lines, so $BC2=BA1$,

for same reason, we have

$CA1=CB2,AC2=AB2$, and $CB2=AC-AB2$.

since $BA1=BC+CA1=BC+CB2=BC+AC-AB2,BC2=AB+AC2=AB+AB2$,

so we have

$BC+AC-AB2=AB+AB2$, ie$AB2=\dfrac{BC+AC-AB}{2}=s-c=CB0, CB2=AC-AB2=AC-CB0=AB0=s-b$ .

with same reason, we have

$AC2=BC0=s-b$,

$BC2=AC0=s-a,BA2=CA0=s-c,BA0=CA2=s-a$.

above facts are for any triangles as we don't have limits for the triangle.

now we check $R1-A2-C-B1$, since $R1A2=R1B1=r1$, so $R1-A2-C-B1$ is a square! we get

$r1=CA2=s-a$,

with same reason , we get

$r2=CB2=s-b$,

we also konw $r3=R3A3=CB3$ ,

since $\angle B3AC2=\angle R3AB3$($AR3$ is bisector), so $AC2=AB3$,

$r3=CB3=AC+AB3=AC+AC2=b+s-b=s$,

clearly:

$r=IA0=CB0=s-c$

now we show the interesting fact:

$r+r1+r2+r3=s-c+s-b+s-a+s=2s=a+b+c$

$r^2+r1^2+r2^2+r3^2=(s-c)^2+(s-b)^2+(s-a)^2+s^2=4S^2+a^2+b^2+c^2-2s(a+b+c)=4S^2+a^2+b^2+c^2-2S*2S=a^2+b^2+c^2$

we rewrite the again:

In Right angle:

$r+r1+r2+r3==a+b+c$

$r^2+r1^2+r2^2+r3^2=a^2+b^2+c^2$

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