Just use two lemma: Lemma 1: Let $\triangle ABC$, $D$ be the point on $(ABC)$. $AD$ cut $BC$ at S. Then $\frac{SB}{SC}=\frac{AB}{AC}. \frac{DB}{DC}.$ Lemma 2: Let $\triangle ABC$, $D$ be the point on $BC$. Then $\frac{DB}{DC}=\frac{\sin DAB}{\sin DAC}.\frac{AB}{AC}$

Call S' is the intersection of AS and BC. Define$ f(X)=\frac{XB}{XC}$ for any point X. By Lemma 1: We have $f(D)=f(D_1)f(A)$ and we need to prove $f(D)=f(S).f(X)$ We got: $f(K)=f(D_1).f(X)$ and $f(S')=f(S).f(A)$ So we need to prove : $f(D)^2 = f(S').f(K)$ By Lemma 2: We have $f(S')=\frac{S'B}{S'C}=\frac{\sin FAT}{\sin EAT} \frac{AB}{AC}=\frac{TF.AE}{TE.AF} \frac{AB}{AC}$ In additon: $f(K)= \frac{\sin KPB}{\sin KPC} \frac{PB}{PC} = \frac{\sin EPT}{\sin FPT} \frac{PB}{PC} =\frac{TE.PF}{TF.PE} \frac{PB}{PC}$ From these two ratio above we have $f(S').f(K)=\frac{PF.PE.AE.AF}{PB.PC.AB.AC}$ Use Lemma 2 for triangle $EPC$ with $A$ on line $EC$ we have: $\frac{AE}{AC}=\frac{\sin APE}{\sin APC} . \frac{PE}{PC} $ Similar we have $\frac{AF}{AB}= \frac{\sin FPA}{\sin BPA} . \frac{PF}{PB} $ So $f(S').f(K)=(\frac{AE}{AC} \frac{AF}{AB})^2 $ Now it is easy, any mistake ?