Let parabola is $y=f(x)=u(x^2-2px+q)$ where $u \neq 0$ Axis of parabola is $x=p$ Let point $A$ is $(a,f(a))$, $B=(b,f(b))$ and $C=(c,f(c))$ Let $M$ is midpoint $BC$ or $M=(\frac{b+c}{2},\frac{f(b)+f(c)}{2})$ And midpoint of $AM$ is $E=(\frac{b+c+2a}{4}, \frac{f(b)+f(c)+2f(a)}{4})$ As bisector of angle $\angle BAC$ is parallel to the axis $Ox$ then it means that $\frac{f(b)-f(a)}{b-a}=-\frac{f(c)-f(a)}{c-a}$ or $a+b-2p=2p-a-c \to 2a+b+c=4p$ But then $E=(p,\frac{f(b)+f(c)+2f(a)}{4})$ so $E$ lies on axis of parabola