• $\angle BIC=90^\circ + \frac{\angle A}{2}$
• $\angle BCO=90^\circ - \angle A$

Since $AC=BC$, hence $\angle A=\angle B$. 

So, 

$\angle BCO=90^\circ - \frac{\angle BOC}{2} $

$=90^\circ - \angle A$

$=\frac{180^\circ - \angle A - \angle A}{2}$

$=\frac{180^\circ - \angle A - \angle B}{2} $

$= \frac{\angle C}{2} $

$= \angle BCI$

Hence $C,O,I$ are collinear, or $CO$ is the angle bisector of $\angle ACB$.

a.

Now, $\angle DIO=\angle DBO=\angle DCO=\angle OCA=\angle ICA$. Therefore, $DI\parallel AC$ or $DI$ does not intersect $AC$.

b. 

Let $K$ be the intersection of $OD$ and $IB$. 

Now,

$\angle BKD$

$=\angle KDC-\angle KBC$

$=\angle ODC-\angle IBC$

$=\angle OIB-\angle B/2$

$=90^\circ + \frac{\angle A}{2} - \frac{\angle B}{2}$

$=90^\circ$