Proof of Theorem 4

(* ================== I N C E N T E R =============== *)

(* Given AB, AC, BC with angle bisecto AD, returns BD, DC *)
FindAngleBisectorDivision[AB_,AC_,BC_]:=Module[{BD,DC},
BD=(AB/(AB+AC))BC;
DC=BC-BD;
Return[{BD,DC}];
];

Clear[K1,K2,K3,K4,K5,K6];
Clear[s1,s2,s3,s4,s5,s6];
Clear[a,b,c];
Clear[AD,BE,CF];
{a1,a2}=FindAngleBisectorDivision[c,b,a];
{b1,b2}=FindAngleBisectorDivision[a,c,b];
{c1,c2}=FindAngleBisectorDivision[b,a,c];
{e,f}=FindAngleBisectorDivision[a,b1,BE]//Simplify;{g,h}=FindAngleBisectorDivision[a,c2,CF]//Simplify;
{j,k}=FindAngleBisectorDivision[a2,b,AD]//Simplify;
eq2=K2/K1==a2/a1;
eq3=(K1+K2)/K3==e/f;
eq4=K4/K3==b2/b1;
eq5=(K3+K4)/K5==g/h;
eq6=K6/K5==c2/c1;
fixK=Solve[{eq2,eq3,eq4,eq5,eq6},{K2,K3,K4,K5,K6}][[1]];
fixs=
{s1->(1/2)(e+a1+j),
s2->(1/2)(g+a2+j),
s3->(1/2)(g+b1+f),
s4->(1/2)(k+b2+f),
s5->(1/2)(c1+h+k),
s6->(1/2)(c2+h+e)};
r1=K1/s1;
r2=K2/s2;
r3=K3/s3;
r4=K4/s4;
r5=K5/s5;
r6=K6/s6;
S=(b-c)(1/r1+1/r2)+(c-a)(1/r3+1/r4)+(a-b)(1/r5+1/r6)

fixK//Factor

{e,f,g,h,k,j}

S2=Simplify[S/.fixK]

num=Collect[Numerator[S2],{s1,s2,s3,s4,s5,s6}]

num2=num/.fixs//Simplify