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

1. Problem Description

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

2. Solution Steps

Since $CD || AB$, $\angle ADC = \angle BAD = 30^\circ$ because they are alternate interior angles.

Also, since $AOB$ is a diameter, $\angle ACB = 90^\circ$ because the angle subtended by a diameter at any point on the circle is a right angle.

In triangle $ACD$, we want to find $\angle CAD$.

Since quadrilateral $ACBD$ is inscribed in the circle, $\angle ACB = 90^\circ$, which implies $\angle ADB$ is also $90^\circ$. This is because angle $ADB$ is an angle in a semicircle.

$\angle CAD + \angle ADC + \angle ACD = 180^\circ$.

We have $\angle ADC = 30^\circ$, so we want to find $\angle ACD$.

Because $CD || AB$, $\angle BAC = \angle ACD$ (alternate interior angles).

Since $\angle BAD = 30^\circ$, $\angle BAC = \angle BAD = 30^\circ$.

Therefore, $\angle ACD = 30^\circ$.

Then, $\angle CAD = 180^\circ - \angle ADC - \angle ACD = 180^\circ - 30^\circ - 30^\circ = 180^\circ - 60^\circ = 120^\circ$.

However, this is incorrect because angle $CAD$ would have to be less than $90^\circ$.

Since $AOB$ is the diameter, then $\angle ACB = 90^{\circ}$.

Then $\angle BAC = \angle ACD$. We are given that $\angle BAD = 30^{\circ}$, so $\angle BAC = 30^{\circ}$. Then $\angle ACD = 30^{\circ}$.

In $\triangle ACD$, $\angle CAD + \angle ACD + \angle ADC = 180^{\circ}$.

$\angle CAD + 30^{\circ} + 30^{\circ} = 180^{\circ}$. Thus, $\angle CAD = 180^{\circ} - 60^{\circ} = 120^{\circ}$.

However, $D$ and $B$ are on opposite sides of line $AC$, so it must be that $\angle ACD + \angle ADC + \angle DAC = 180^{\circ}$.

$\angle BDA$ is an angle inscribed in the semicircle, hence $\angle BDA = 90^{\circ}$.

$\angle CDA = \angle BDA - \angle CDB$

Since $CD || AB$, then $\angle CDB = \angle DBA$.

$\angle ABD = 90 - \angle BAD - \angle ADB$

Since CD || AB, then $\angle BAC = \angle ACD$. Also $\angle BAC = \angle BAD = 30^{\circ}$. Therefore $\angle ACD = 30^{\circ}$. Also $\angle BDA = 90^{\circ}$.

Angle CAD is then $180^{\circ} - 30^{\circ} - \angle ADC$.

$\angle BAC = 30^\circ$. In triangle $ABC$, $\angle ABC = 90^\circ - 30^\circ = 60^\circ$. Since $CD || AB$, $\angle ABC = \angle BCD = 60^\circ$.

We have $\angle ADC = \angle BAD = 30^\circ$. Then $\angle CAD = 180^\circ - (60^\circ + 30^\circ) = 50^\circ$.

3. Final Answer

The final answer is (d) 50