This Notebook contains supplementary material related to the paper
Stanley Rabinowitz, "Relationships Between Six Excircles", Sangaku Journal of Mathematics, 3 (2019)73-90
which can be found at http://www.sangaku-journal.eu/2019/SJM_2019_73-90_Rabinowitz.pdf.

This Notebook is Copyright © September 2019 by Stanley Rabinowitz (stan.rabinowitz@comcast.net) and is published with open access. This Notebook is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author and the source are credited.
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Proof of Theorem 5.1
From the paper, it suffices to prove Lemma 5.2.

(* Given a, b, d, e, this returns t *)
Stewart[a_,b_,d_,e_]:=Sqrt[(a^2e+b^2d-(d+e) d e)/(d+e)];

(* Use Stewart's Theorem to find AD, BE, CF *)

AD=Stewart[c,b,s-c,s-b]//Simplify

BE=Stewart[a,c,s-a,s-c]//Simplify

CF=Stewart[b,a,s-b,s-a]//Simplify

(* Show that these agree with the formulas given in the paper *)

(* Apply Heron's Formula *)

K=Sqrt[s(s-a)(s-b)(s-c)]

ADpaper=Sqrt[s^2-4K^2/(a(s-a))]

Simplify[(AD-ADpaper)/.s->(a+b+c)/2]

BEpaper=Sqrt[s^2-4K^2/(b(s-b))]

Simplify[(BE-BEpaper)/.s->(a+b+c)/2]

CFpaper=Sqrt[s^2-4K^2/(c(s-c))]

Simplify[(CF-CFpaper)/.s->(a+b+c)/2]

(* From the paper, we have: *)

Proof of Lemma 5.2

AN=AD (a/s);
ND=AD (s-a)/s;
BN=BE (b/s);
NE=BE (s-b)/s;
CN=CF (c/s);
NF=CF(s-c)/s;
s1=((s-c)+ND+BN)/2;
s2=((s-b)+CN+ND)/2;
s3=((s-a)+CN+NE)/2;
s4=((s-c)+AN+NE)/2;
s5=((s-b)+AN+NF)/2;
s6=((s-a)+BN+NF)/2;
expr=s1 s3 s5-s2 s4 s6/.{s->(a+b+c)/2};
Simplify[expr]