In the given circle, if angle ACB = 30 degrees, find angles BOC, BAC, and BDC.

Theorem: Angle in a semicircle is $90^\circ$.
Since AB is a diameter, the angle $\angle ACB$ subtended by the diameter at point C on the circumference is $90^\circ$.
Theorem: Sum of angles in a triangle is $180^\circ$.
In $\triangle ABC$, the sum of its interior angles is $180^\circ$. We have $\angle CAB = 18^\circ$, $\angle ACB = 90^\circ$, and $\angle CBA = a$.

$$18^\circ + a + 90^\circ = 180^\circ$$ $$a + 108^\circ = 180^\circ$$ $$a = 180^\circ - 108^\circ$$ $$a = 72^\circ$$

Theorem: Angles subtended by the same arc at the circumference are equal.
The angle $b$ is $\angle CAD$. This angle subtends arc CD.
The angle $\angle CBD$ also subtends arc CD. From the diagram, $\angle CBD = 50^\circ$.
Since both angles subtend the same arc, they are equal.

$$b = \angle CBD$$ $$b = 50^\circ$$

Theorem: Opposite angles in a cyclic quadrilateral sum to $180^\circ$.
ABCD is a cyclic quadrilateral because all its vertices lie on the circle.
The angle $c$ is $\angle BCD$. Its opposite angle in the cyclic quadrilateral is $\angle DAB$.
Therefore, $\angle DAB + \angle BCD = 180^\circ$.
First, calculate the measure of $\angle DAB$:

$$\angle DAB = \angle DAC + \angle CAB$$ $$\angle DAB = b + 18^\circ$$

Substitute the value of $b$ from Step 2:

$$\angle DAB = 50^\circ + 18^\circ$$ $$\angle DAB = 68^\circ$$

Now, substitute this value into the cyclic quadrilateral property:

$$68^\circ + c = 180^\circ$$ $$c = 180^\circ - 68^\circ$$ $$c = 112^\circ$$

Final values:
$a = \boxed{72^\circ}$
$b = \boxed{50^\circ}$
$c = \boxed{112^\circ}$