In the given figure, $AOB$ is a diameter of a circle, and $CD \parallel AB$. If $\angle BAD = 30^\circ$, we need to find the measure of $\angle CAD$.

1. Problem Description

In the given figure, $AOB$ is a diameter of a circle, and $CD \parallel AB$. If $\angle BAD = 30^\circ$, we need to find the measure of $\angle CAD$.

2. Solution Steps

Since $CD \parallel AB$, we know that $\angle CDA = \angle DAB$ because they are alternate interior angles. Therefore, $\angle CDA = 30^\circ$.

The angle subtended by the diameter at the circumference is a right angle. Thus, $\angle ACB = 90^\circ$.

In triangle $ABC$, $\angle CAB + \angle ABC + \angle ACB = 180^\circ$.

Since $CD \parallel AB$, $\angle DCA = \angle CAB$ are alternate interior angles.

Since $CD \parallel AB$, $\angle CDB = \angle DBA$ are alternate interior angles.

However, angles subtended by the same chord on the circumference are equal, so $\angle DCA = \angle DBA$, and thus $\angle CAB = \angle CDB$.

But $\angle CDB = \angle CDA = 30^{\circ}$.

Then $\angle CAB = 30^{\circ}$.

Consider the triangle $ACD$.

$\angle CAD + \angle ADC + \angle DCA = 180^{\circ}$

We know that $\angle ADC = 30^{\circ}$, and $\angle DCA = \angle CAB$.

We also know that $\angle CAB = 30^{\circ}$.

Therefore $\angle DCA = 30^{\circ}$.

Thus, $\angle CAD + 30^{\circ} + 30^{\circ} = 180^{\circ}$ does not seem relevant for this question.

Since $AOB$ is a diameter, $\angle ACB = 90^{\circ}$. In triangle $ABC$, $\angle CAB + \angle ABC = 90^{\circ}$.

$\angle CAB = \angle DCA$ are alternate interior angles since $CD \parallel AB$.

$\angle DCA = \angle DBA$ because angles in the same segment subtended by chord $AD$ are equal.

Thus, $\angle DBA = \angle DCA$.

We also have $\angle BAD = 30^{\circ}$.

Consider quadrilateral $ACDB$. This is a cyclic quadrilateral.

In a cyclic quadrilateral, opposite angles are supplementary. Therefore $\angle ACB + \angle ADB = 180^{\circ}$.

Thus $90^{\circ} + \angle ADB = 180^{\circ}$. $\angle ADB$ must then be $90^\circ$.

Since $\angle ADB = \angle ADC = 30^\circ + \angle CDB$, this gives us $90^{\circ} = 30^{\circ} + \angle CDB$,

Therefore $\angle CDB = 60^\circ$.

Since $\angle CAB = \angle CDB$ , we have $\angle CAB = 60^\circ$.

Finally, $\angle CAD = \angle CAB - \angle DAB = 60^\circ - 30^\circ = 30^\circ$.

3. Final Answer

30 degrees