Symbolically solve the equilateral case
In[]:=
a={1,Sqrt[3]};b={0,0};c={2,0};
In[]:=
(*inradiusofAECintermsofλ1=BE/BA*)​​expr1=FullSimplify[TriangleMeasurement[{λ1*(a-b)+b,a,c},"Inradius"],Assumptions0<λ1<1]
Out[]=
(-1+λ1)-2+λ1+
1+(-1+λ1)λ1

2-λ1+
1+(-1+λ1)λ1
In[]:=
(*FindthelocationofPbasedofλ1=BE/BA=λ2=CD/CAfromsymmetry*)​​p=segmentIntersection[b,λ1*(a-c)+c,c,λ1*(a-b)+b][[-1]]
Out[]=
-
2-λ1
-2+λ1
,-
3
λ1
-2+λ1

In[]:=
(*inradiusofPBC*)​​expr3=FullSimplify[TriangleMeasurement[{p,b,c},"Inradius"],Assumptions0<λ1<1]
Out[]=
-2+λ1+2
1+(-1+λ1)λ1
2-λ1+2
1+(-1+λ1)λ1
In[]:=
(*inradiusofPBC=inradiusofAEC*)​​sol=Solve[{expr1expr3,1>λ1>0},λ1]
Out[]=
λ1
2
5
(-1+
6
)
In[]:=
(*findtheinradiusbasedonλ1*)​​expr1/.sol[[1]]
Out[]=
-1+
2
5
(-1+
6
)-2+
2
5
(-1+
6
)+
1+
2
5
(-1+
6
)-1+
2
5
(-1+
6
)
2-
2
5
(-1+
6
)+
1+
2
5
(-1+
6
)-1+
2
5
(-1+
6
)
In[]:=
(*simplifyanddenest*)​​FullSimplify[%]
Out[]=
5-2
6
In[]:=
ResourceFunction["RadicalDenest"][%]
Out[]=
-
2
+
3
(*sameastheresultfrommynumericmethod*)
In[]:=
N[%]
Out[]=
0.317837