Let $BI$ interect $(ABC)$ at $U$. By Incenter Lemma, we get that $UA=UC=UI$. Using Th. Ptolemeu in $BAUC$, we get $AB*UC+BC*UA=BU*AC$. So, $IU(AB+BC)=BU*AC$. We will prove that $AB+BC=3AC$, by proving $3IU=BU$. Let $T$ be a point on $AU$ such that $U$ is the midpoint of $(IT)$. Again, from Incenter Excenter Lemma, we get that T is the B-excenter of triangle $ABC$. So, $\angle IAT=\angle ICT=90$. But, we know $\angle DIA=90$, so $DI || AT$, thus $I$ is the midpoint of $BT$. So, $3IU=UB$. Thus, $AB+BC=3AC$ and we are done.