Sketch: Call projections P and Q. Use IRAN lemma then prove that this circle is the nine-point-circle of the triangle formed by the lines AP, AQ and BC. In other words, tangency point is one of the midpoints of new triangle.

Let the incircle with center $I$ touch $BC$, $CA$, and $AB$ at $D$, $E$, $F$ respectively. Let the projections from $A$ to $BI$, $CI$, and $BC$ respectively be $X$, $Y$, and $Z$. By the Iran Lemma, $X$ lies on $DE$ and $Y$ lies on $FD$. Then notice that $\angle XDY = 90^\circ - \frac{\angle A}{2}$ and the quadrilaterals $AXZB$ and $AYZC$ are cyclic. Thus, we see that $\angle XZY = \angle XZA + \angle YZA = \angle XBA + \angle YCA = \angle IBA + \angle ICA = 90^\circ - \frac{\angle A}{2}$. Therefore since $\angle XZY = \angle XDY$, we see that $XDZY$ is cyclic. $\blacksquare$