Let $H$ be the orthocenter of the triangle $ABC$. Let the points $K, M, L$ be the intersection of the lines $AI$ and $OD$, $AI$ and $OH$, $AH$ and $OI$. We know that $AH = 2OD$. By a simple angle chasing we see that $AI$ is an angle bisector of $HAO$. Therefore
$$\frac{AO}{OD} = 2\frac{AO}{AH} = 2\frac{OM}{MH}$$
(1).

That is easy to see that $K$ is on the circumcircle of $ABC$. Let $\angle A = 2\alpha$. Then we have $BC = 2R\sin(2\alpha)$ and $AT = r \cdot \cot\alpha$. Then
$$\frac{BC}{AT} = 2\frac{2R\sin\alpha}{r/\sin\alpha} = 2\frac{BK}{AI} = 2\frac{KI}{AI} = 2\frac{OI}{IL}$$
(2) since $BK = KI$ and $AH \parallel OK$.

Let $HM = x$, $MP = y$, $IL = k$, $IQ = s$. Then $PO = x + y$ by the very well known fact that $P$ is the midpoint of $[HO]$, $OQ = s$, $OM = x + 2y$, $OI = 2s$, $QL = k + s$. Applying Menelaus Theorem on the triangle $ALI$ with respect to the collinear points $H, M, O$ gives
$$\frac{AH}{AL} = \frac{OIHM}{ILMO} = \frac{2sx}{k(x+2y)}.$$
Applying Menelaus Theorem on the triangle $ALQ$ with respect to the collinear points $H, P, O$ gives
$$\frac{AH}{AL} = \frac{OQHP}{QLPO} = \frac{s}{k+s}.$$
Therefore $2x(k+s) = k(x+2y)$, that is
$$\frac{y}{x} - \frac{s}{k} = \frac{1}{2}$$
(3).

Finally (1), (2) and (3) result
$$\frac{AO}{OD} - \frac{BC}{AT} = 2\frac{x+2y}{x} - 2\frac{2s}{k} = 2+4\left(\frac{y}{x} - \frac{s}{k}\right) = 2+2=4.$$