Let $D,E$ denote the midpoints of the arcs $BC,CA$ of the circumcircle $\odot(ABC)$ and $J \equiv DB_1 \cap CC_1.$ $\widehat{A}:\widehat{B}:\widehat{C}=4:2:1$ $\Longrightarrow$ $\widehat{A}=2 \cdot \widehat{B}$ and $\widehat{B}= 2 \cdot \widehat{C}$ $\Longrightarrow$ $\triangle BCB_1$ is isosceles with apex $B_1$ $\Longrightarrow$ $BCEA$ is isosceles trapezoid with bases $AE,BC,$ thus by symmetry, we have $\widehat{JDA_1}=\tfrac{1}{2}\widehat{ABE}=\widehat{JCA_1}$ $\Longrightarrow$ $DCJA_1$ is cyclic $\Longrightarrow$ $\widehat{DJA_1}=\widehat{DCB}=\widehat{CAD}$ $\Longrightarrow$ $AA_1JB_1$ is cyclic $\Longrightarrow$ $\widehat{AA_1J}=\widehat{DCJ}=\tfrac{1}{2}\widehat{C}+\tfrac{1}{2}\widehat{A}=\tfrac{1}{2}\widehat{C}+\widehat{B}=\widehat{AC_1J}$ $\Longrightarrow$ $AJA_1C_1$ is cyclic $\Longrightarrow$ $AB_1A_1C_1$ is cyclic and since $AA_1$ bisects $\angle B_1AC_1,$ then $A_1B_1=A_1C_1.$