在图中，A、B、C 和 D 是以 O 为圆心的圆周上的点。AB 平行于 DO 且 $\angle AOD = 67^\circ$。AOC 是该圆的直径。

1. Find the base angle in isosceles triangle $AOD$.

Since $OA=OD$ as radii of the circle, $\triangle AOD$ is isosceles.

$\angle OAD=\angle ODA=\frac{180^\circ-67^\circ}{2}=56.5^\circ$

2. Relate this angle to the exterior angle at $A$ using parallel lines.

Since $AB \parallel DO$, the angle between the extension of $AB$ and $AD$ is equal to $\angle ODA$.

So the exterior angle at $A$ of cyclic quadrilateral $ABCD$ is

$56.5^\circ$

3. Use the cyclic quadrilateral angle property to find $\angle BCD$.

In a cyclic quadrilateral, an exterior angle is equal to the interior opposite angle.

Therefore,

$\angle BCD=56.5^\circ$

1. Use the earlier result to identify the supplementary angle at A.

From part (a), $\angle BCD = 56.5^\circ$.

Since $A$, $B$, $C$ and $D$ are points on the circumference, angles in the same segment are equal, so
$\angle BAC = \angle BCD = 56.5^\circ.$

2. Use the diameter line to explain why angle DAB is supplementary to angle BAC.

Since $AOC$ is a diameter, $A$, $O$ and $C$ are collinear, so $AC$ is a straight line through $A$.

Hence
$\angle DAB = 180^\circ - \angle BAC = 180^\circ - 56.5^\circ = 123.5^\circ.$

1. Identify the angle at A subtended by chord $DC$.

Since $A$, $O$ and $C$ are collinear, $AC$ is the same straight line as $AO$.

From the earlier working, $\angle OAD = 56.5^\circ$.

Hence
$\angle DAC = 56.5^\circ.$

2. Use angles in the same segment to find $\angle DBC$.

$\angle DAC$ and $\angle DBC$ are angles in the same segment, subtended by the same chord $DC$.

Therefore
$\angle DBC = \angle DAC = 56.5^\circ.$