I assume that you mean non-isosceles triangle $ABC$. Pick any line $l$ parallel to $BC$ and let it intersect $HB_1$, $HC_1$ and $HA$ at $X$, $Y$ and $Z$. We have $l\perp HA$ and since $HA$ is the angle bisector of $B_1HC_1$ we have $|XZ|=|YZ| \Rightarrow (X,Y;Z,\infty)=-1$. Thus $-1=(X,Y;Z,\infty)\overset{H}{=}(B_1,C_1;HA\cap B_1C_1,BC\cap B_1C_1)\overset{A}{=}(B_1,C_1,H,BC\cap B_1C_1)$. Hence, it is clear that $B_1C_1$ always pass through a fixed point on $BC$. Now, let $D$ and $E$ be the feet from $B$ and $C$. Let $P=DE\cap BC$ and remeber that $B_1C_1$ always pass through $P$. If $BCB_1C_1$ is cyclic then by Reim's theorem we have $B_1C_1||DE$. But since $P=B_1C_1\cap DE$ this means that $P=BC_{\infty}$. To conclude notice that $-1=(P,H;B,C)=(BC_{\infty},H;B,C)\Rightarrow |BH|=|HC| \Rightarrow |AB|=|AC|$.