Let $H$ be the orthocenter of $\triangle ABC.$ $P$ is midpoint of $\overline{OH}.$ $AI$ cuts the circumcircle $(O)$ of $\triangle ABC$ again at midpoint $M$ of its arc $BC.$ Since $AH=2 \cdot OD,$ then the reflection $S$ of $O$ about $BC$ forms the parallelogram $AHSO$ $\Longrightarrow$ $AS,OH$ bisect each other at $P.$ Hence if $A,P,Q$ are collinear, then $A,Q,S$ are collinear. By Menelaus theorem for $\triangle OMI,$ cut by $\overline{AQS},$ we get $\frac{AI}{AM} \cdot \frac{MS}{SO} \cdot \frac{OQ}{QI}=1 \Longrightarrow \frac{AM}{AI}=\frac{MS}{SO}=\frac{AO-2 \cdot OD}{2 \cdot OD} \Longrightarrow$ $\frac{AO}{OD}=2 \cdot \frac{AM}{AI}+2=\left(1+\frac{p}{p-a} \right)+2 =4+\frac{a}{p-a}=4+\frac{BC}{AT}.$

hello; thank you, Luis Gonzalez. (the beautiful solution ). But i have an other solution, it is very nice too. Let $H$ - orthocenter, $AA_1,BB_1,CC_1$ - altitudes, $r,R$ - inradius and circumradius of the triangle $ABC$. Idea: $A,P,Q$ are collinear if and only if \[ \frac{HB_1+OE}{HC_1+OF} = \frac{OE+r}{OF+r} \] We have very nice two lemmas: (1) $HA_1 \cdot R=2 d_b \cdot d_c $ and ... (2) $d_a+d_b+d_c = r+R $. So, since its, we see not difficult \[ \frac{AO}{OD} - \frac{BC}{AT}=4 .\]

Since it's not here, I'd like to note that a straightforward bary calculation is very possible. First, we calculate using trivial identities as follows.$$\frac{AO}{OD} - \frac{BC}{AT} = 4 \iff \sec A - \frac{2a}{b+c-a} = 4 \iff \frac{b^2+c^2-a^2}{2bc} = \frac{b+c-a}{4b+4c-2a} \iff (b^2+c^2-a^2)(2b+2c-a) = bc(b+c-a)$$ Now, since $\vec {GP} = -\frac{1}{2} \vec{GO}$ and $\vec {Q} = \frac{1}{2} (\vec{O} + \vec{I})$, it is straightforward to calculate the coordinates. Denote $F_a = a^2(b^2+c^2-a^2)$, $F_b = b^2(c^2+a^2-b^2)$, $F_c = c^2(a^2+b^2-c^2)$. After some calculations using the well-known coordinates for $O, G, I$, we will end up with $$P(F_b+F_c : F_c+F_a : F_a+F_b)$$$$Q(F_a (a+b+c) + a(F_a+F_b+F_c) : F_b(a+b+c) + b(F_a+F_b+F_c) : F_c(a+b+c) + c(F_a+F_b+F_c))$$ So we have $APQ$ colinear iff $\frac{F_a+F_b}{F_a+F_c} = \frac{F_c(a+b+c) + c(F_a+F_b+F_c)}{F_b(a+b+c)+b(F_a+F_b+F_c)}$. This is equivalent to $(F_b - F_c)(F_a+F_b+F_c)(a+b+c) = (F_a+F_b+F_c) ((c-b)F_a + cF_c - bF_b)$. It suffices to work with $(F_b-F_c) (a+b+c) - (c-b)F_a - cF_c + bF_b =0$. The L.H.S factorizes as $(c-b)(a+b+c)( (b^2+c^2-a^2)(2b+2c-a) - bc(b+c-a)) = 0$, so we're done.