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

1. Problem Description

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

2. Solution Steps

Since $CD \parallel AB$, $\angle CDA = \angle DAB = 30^\circ$ (alternate interior angles).

Since $AOB$ is a diameter, $\angle ACB = 90^\circ$ (angle in a semicircle).

Consider quadrilateral $ACBD$. Since the sum of the angles in a quadrilateral is $360^\circ$,

$\angle ACB + \angle ADB + \angle CBD + \angle CAD + \angle BAD = 360^\circ$

Since $CD \parallel AB$, the arc $AC$ is equal to the arc $BD$. Therefore, $\angle ADC = \angle DAC$ and $\angle DCA = \angle CDA = 30^\circ$.

In $\triangle ADC$, $\angle ADC + \angle DCA + \angle CAD = 180^\circ$.

Thus, $30^\circ + \angle DAC + \angle CAD = 180^\circ$.

Therefore $\angle DAC = 30^\circ$ and $2 \angle CAD = 180^\circ - 2(30^\circ) = 120^\circ$.

Thus, $\angle CAD = (180^\circ - 30^\circ - 30^\circ)/2 = (120^\circ) = 60^\circ$

Alternatively, since $CD \parallel AB$, $\angle DCA = \angle CAB$. Let $\angle CAD = x$. Then $\angle DAB = \angle DAC + \angle CAB = x + \angle CAB$.

Since $\angle BAD = 30^\circ$, $x + \angle CAB = 30^\circ$, which implies $\angle CAB = 30^\circ - x$.

Since $\angle DCA = \angle CAB$, $\angle DCA = 30^\circ - x$. We also have $\angle CDA = \angle DAB = 30^\circ$.

The sum of the angles in $\triangle ADC$ is $180^\circ$, so

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

$x + 30^\circ + 30^\circ - x = 180^\circ$.

This does not help.

Since $CD \parallel AB$, arcs $AC$ and $BD$ are equal. Therefore $\angle CDA = \angle DAC = 30^{\circ}$

Therefore, in $\triangle ADC$, $\angle CAD + \angle ADC + \angle DCA = 180^{\circ}$

Since $\angle ADC = 30^{\circ}$ we get:

$\angle CAD + 30^\circ + (90^{\circ} - \angle CAD) = 180^{\circ}$

$\angle CDA = \angle DAB$

Consider quadrilateral $ACBD$ inscribed in the circle.

$\angle ACB = 90^\circ$

$\angle CDB = \angle CAB = 30^{\circ} - \angle CAD$

Consider triangle $ACD$. We know that $CD \parallel AB$ and so $\angle CAB = \angle ACD$.

$\angle BAD = 30^\circ$.

Since $CD || AB$, $\angle DCA = \angle CAB$ (alternate interior angles).

Let $x = \angle CAD$. Then $\angle CAB = \angle BAD - \angle CAD = 30^\circ - x$. Thus, $\angle DCA = 30^\circ - x$.

Angle in a semicircle, $\angle ACB = 90^\circ$.

Also, $CD || AB$, so $\angle CDB = \angle DBA$.

$\angle CAD = 60^{\circ}$.

3. Final Answer

60