Step by step solution:

With $\widehat{CAx}$ is the outer angle of $ΔABC$

$\Rightarrow \widehat{CAx} = 180^o - \widehat{CAB} = 180^o - 120^o = 60^o$

We have:

$AA_1$ is the bisector of $\widehat{BAC}$

$\Rightarrow \widehat{BAA_1} = \widehat{CAA_1} = \dfrac{1}{2}\widehat{BAC} = 60^o$

$\Rightarrow \widehat{CAA_1} = \widehat{CAx} = 60^o$

$\Rightarrow AC$ is the bisector of $\widehat{A_1Ax}$

$\Rightarrow AC$ is the bisector of the outer angle $\widehat{A_1Ax}$ of $ΔABA_1$

Beside, $BB_1$ is the bisector of $\widehat{ABA_1}$

and $B_1$ is the intersection point of $AC$ and $BB_1$

So $B_1$ is the center of the escribed circle  tangent to side $AA_1$ of $ΔABA_1$

Therefore: $A_1B_1$ is the bisector of the outer angle $\widehat{AA_1C}$ of $ΔABA_1$

$\Rightarrow \widehat{AA_1B_1} = \dfrac{1}{2}\widehat{AA_1C}$

By proving the same way as above, we get:

$A_1C_1$ is the bisector of the outer angle $\widehat{AA_1B}$ of $ΔACA_1$

$\Rightarrow \widehat{AA_1C_1} = \dfrac{1}{2}\widehat{AA_1B}$

Hence:

$\widehat{B_1A_1C_1} = \widehat{AA_1B_1} + \widehat{AA_1C_1} = \dfrac{1}{2}(\widehat{AA_1C} + \widehat{AA_1B}) = 90^o$

So that $A_1B_1C_1$ is a right angle