any solution

A formula that we'll use a lot is this: $ XI = Y\frac{\sin \frac {1}{2} Z}{\cos \frac {1}{2} Y} = Z\frac{\sin \frac {1}{2} Y}{\cos \frac{1}{2} Z}$, where I is the incenter of $\Delta XYZ$. By that, and by noticing that $\Delta A_{1}B_{1}C_{1}$ is the orthic triangle of $\Delta ABC$, we find that $J_{y}X_{1} = X_{1}Y \frac {\sin \frac{1}{2} Y}{\cos \frac{1}{2} Z} = XI \cos Y$, using XYZ as any permutation of ABC. Using the law of sines on $\Delta J_{y}X_{1}J_{z}$, we have that $\frac {\cos Y}{\sin \measuredangle X_{1}J_{z}J_{y}} = \frac {\cos Z}{\sin \measuredangle X_{1}J_{y}J_{z}}$. We know that $\measuredangle X_{1}J_{z}J_{y} + \measuredangle X_{1}J_{y}J_{z} = X$ So we can write $\frac{\sin (\frac {\pi}{2} - Y)}{\sin (\frac {\pi}{2} - Z)} = \frac{\sin \measuredangle X_{1}J_{z}J_{y}}{\sin \measuredangle X_{1}J_{y}J_{z}}$ and the sum of the angles on both sides is X. Using a lemma, which I will only state here, $\measuredangle X_{1}J_{z}J_{y} = \frac{\pi}{2} - Y$ and $\measuredangle X_{1}J_{y}J_{z} = \frac{\pi}{2} - Z$. We have, doing this to every angle and noticing that $\measuredangle Z_{1}J_{x}Y_{1} = \frac{\pi + A}{2} $, that $\measuredangle J_{y}J_{x}J_{z} = \frac{\pi - A}{2}$. Law of sines again on $\Delta J_{y}X_{1}J_{z}$, we find that $J_{y}J_{z} = XI \sin X $. Law of sines on $\Delta J_{x}J_{y}J_{z}$, $R_{J} = XI \sin \frac {1}{2} X = 4R \sin \frac {1}{2} X \sin \frac {1}{2} Y \sin \frac{1}{2} Z = 4R \sin \frac {1}{2} A \sin \frac {1}{2} B \sin \frac{1}{2} C = r$, where r is the inradius and R is the circumradius. Q.E.D

Note that $AC_1B_1$ and $A_1BC_1$ are similar so $J_bC_1J_a$ and $BC_1B_1$ are similar also $J_aB_1J_c$ and $CB_1C_1$ are similar so $\angle J_bJ_aJ_c = \frac{\angle B+C}{2}$. By sines law in $A_1J_bJ_C$ we have $\frac{\sin{A}}{J_bJ_c} = \frac{\sin{90-B}}{A_1J_b}$ Note that $A_1J_b = AI.\cos{B}$ so $J_bJ_c = AI.\sin{A}$. Let $R$ be circumradius of $J_aJ_bJ_c$. Note that $2R = \frac{AI.\sin{A}}{\sin{\frac{B+C}{2}}}$ and $2r = (b+c-a).\sin{\frac{A}{2}}$ so we need to prove $\frac{AI.\sin{A}}{\sin{\frac{B+C}{2}}} = (b+c-a).\sin{\frac{A}{2}}$ or $\frac{\sin{\frac{A}{2}}}{\sin{A}} = \frac{AI}{b+c-a}$ which is true. we're Done.