Denoting by $c_{1}$ and $c_{2}$ the two incircles centered at $I_{1},I_{2}$, introduce points $R_{1},R_{2}$ the points of tangency from $X$ to $c_{1},c_{2}$ respectively and $S_1,S_2$ the points where $c_{1},c_{2}$ are respectively tangent to $AP$ . Moreover, denote by $T_1,T_2$ the to contact points of the two circles with $BC$ Note that $\angle R_1XR_2 =\angle R_1XP+\angle PXR_2= 2 \cdot (\angle I_1XP+\angle PXI_2)=2 \cdot 90 =180 $ So $R_1R_2$ is a common tangent of $c_1,c_2$ Also reflecting over $I_1I_2$ , $R_1R_2=T_1T_2$ Now, $R_1R_2=R_1X+XR_2=XS_1+XS_2=S_1S_2+2 \cdot XS_2$ and $T_1T_2=T_1P+PT_2=PS_1+PS_2=S_1S_2+2 \cdot PS_1 \implies XS_2=PS_1$ we have $AX=AS_2-XS_2=AS_2-PS_!$ But $AX=AS_2-PS_1=\frac{AB+AP-BP}{2}-\frac{AP+PC-AC}{2}=\frac{AB+AC-BC}{2}$ which is evidently indepedent of $P$

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