Let incircle of $\triangle ABC$ touch $BC,CA$ at $E,F, EF\cap BC\equiv N'$. Since $(B,C;D,N')=-1, \frac{BD}{CD}=\frac{BN'}{CN'}$, remains to show $N'\equiv N$. Note $\triangle DEF\sim \triangle IPQ$, with angles $90^{\circ}-\frac{\angle A}{2}, 90^{\circ}-\frac{\angle B}{2}, 90^{\circ}-\frac{\angle C}{2}$. Let $M$ be the midpoint of $EF$, then $A,M,H,N'$ lie on a circle with diameter $AN'\implies \angle IAP=\angle END$. Thus $\triangle DEF\cup N'\cup I\sim \triangle IPQ\cup A\cup O\implies \angle IAO=\angle IN'D\implies A,I,L,N'$ concylic$\implies N\equiv N'$.

buzzychaoz wrote: In acute triangle $ABC, AB<AC$, $I$ is its incenter, $D$ is the foot of perpendicular from $I$ to $BC$, altitude $AH$ meets $BI,CI$ at $P,Q$ respectively. Let $O$ be the circumcenter of $\triangle IPQ$, extend $AO$ to meet $BC$ at $L$. Circumcircle of $\triangle AIL$ meets $BC$ again at $N$. Prove that $\frac{BD}{CD}=\frac{BN}{CN}$. Let $\triangle DEF$ be the contact triangle of $\triangle ABC$. Redefine $N$ as the point on line $BC$ with $(B,C;D,N)=-1$. Observe that $N$ lies on line $EF$. Note $\triangle IPQ \sim \triangle DEF$ and consider the spiral similarity $\mathcal{S}: \triangle IPQ \mapsto \triangle DEF$. Since $IA$ is tangent to $\odot(IPQ)$ and $DN$ is tangent to $\odot(DEF)$ we conclude that $A \mapsto N$. Finally, $O \mapsto I$. Now $\triangle OIA \sim \triangle IDN$ hence $\angle IAL=\angle IAO=\angle IND$ proving $AILN$ is cyclic as desired.

A slightly different solution: Similarly, redefine $N$ as the unique point with $(B, C; D, N)=-1$ and let $E$, $F$ be the point of tangency on the sides $AC$ and $AB$. It's well-known that $N-E-F$ and $AD \parallel IN$. Now suppose $M$ is the midpoint of $PQ$, we have $\angle{MOI}=2\angle{QPI}+\angle{QIP}=\pi-\angle{B}+\frac{\pi}{2}-\frac{\angle{A}}{2}$ and $\angle{MAI}=\frac{\angle{A}}{2}-\angle{BAH}=\frac{\angle{A}}{2}+\angle{B}-\frac{\pi}{2}$, so $\angle{MOI}+\angle{MAI}=\pi$, implying that $A, M, O, I$ are concyclic. Looking at $\triangle{MAI}$, since its circumcenter and orthocenter are isogonal conjugates (and $AO$ is the diameter of $(AMOI)$), $\angle{DAI}=\angle{MAO}$, meaning that we have $\angle{AIN}=\angle{ALN}$, so $A, I, L, N$ are concyclic as desired.