$\triangle ABC$ is an isosceles triangle with $AC=BC$ and $\angle ACB<60^\circ$. $I$ and $O$ are the incenter and circumcenter of $\triangle ABC$. The circumcircle of $\triangle BIO$ intersects $BC$ at $D \neq B$.
(a) Do the lines $AC$ and $DI$ intersect? Give a proof.
(b) What is the angle of intersection between the lines $OD$ and $IB$
Source: BdMO 2016 National Secondary P6
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,
$\angle BKD$
$=\angle KDC-\angle KBC$
$=\angle ODC-\angle IBC$
$=\angle OIB-\angle B/2$
$=90^\circ + \frac{\angle A}{2} - \frac{\angle B}{2}$
$=90^\circ$
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