, 2015

Let $I$ denote the incenter of $ABC$, i.e. the intersection of the bisectors $l_A$, $l_B$ and $l_C$.

We denote by $P$ and $Q$ the bases of the perpendiculars from $A$ to $l_C$ and $C$ to $l_A$, respectively. Let $PQ$ intersect $AB$ and $BC$ at $R$ and $S$ respectively. Since $APQC$ is inscribed, we have $\angle RPI = \angle QPC = \angle QAC = \angle RAI$, thus $APRI$ is also inscribed. Hence $R$ is the base of the perpendicular from $I$ to $AB$. Similarly for $S$. Since $I$ is the incenter, it follows that $IB \perp RS$ and thus $l_B \perp PQ$. Hence it suffices to prove that $MB_0 \perp PQ$.

By Thales theorem, the point $B_0$ is the circumcenter of $APQC$. Thus
$$B_0P = B_0Q. \qquad (1)$$
Now let $U$ and $V$ denote the bases of the perpendiculars from $B$ to $l_C$ and $l_A$ respectively. We claim that $M$ is the incenter of $PUVQ$. Firstly, since
$$\angle VUI = \angle VBI = \angle VQP,$$
the quadrilateral $PUVQ$ is inscribed. Secondly, since $AP \parallel MC_0 \parallel BU$ and $C_0$ is the midpoint of $AB$, we see that $MP = MU$. Similarly, $MV = MQ$. Consequently $M$ is the circumcenter of $PUVQ$ and therefore $MP = MQ$. Combining with (1), we see that $MB_0 \perp PQ$.