quadrilateral abcd is inscribed in circle o below. segment $\overline{ac}$ is a diameter of the circle.
what is the sum of the degree measures of $\angle abc$ and $\angle cda$?
$\circ$ $180^\circ$
$\circ$ $360^\circ$
$\circ$ $270^\circ$
$\circ$ $90^\circ$

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# Brief Explanations:
1. Recall the property of a cyclic quadrilateral: Opposite angles in a cyclic quadrilateral sum to $180^\circ$. Also, an angle inscribed in a semicircle is a right angle? Wait, no, wait. Wait, $AC$ is a diameter. So $\angle ABC$ and $\angle CDA$: Wait, quadrilateral $ABCD$ is cyclic (inscribed in a circle). Also, since $AC$ is a diameter, $\angle ABC$ and $\angle ADC$ (wait, the question has $\angle CDA$, which is the same as $\angle ADC$): Wait, actually, in a cyclic quadrilateral, opposite angles sum to $180^\circ$? Wait, no, wait. Wait, no, a cyclic quadrilateral's opposite angles are supplementary (sum to $180^\circ$)? Wait, no, wait, no—wait, no, the sum of all interior angles of a quadrilateral is $360^\circ$, and in a cyclic quadrilateral, opposite angles are supplementary. Wait, but here, $AC$ is a diameter. So $\angle ABC$ and $\angle ADC$ (since $AC$ is diameter, $\angle ABC$ is an angle inscribed in a semicircle? Wait, no, $\angle ABC$ is inscribed in the circle with $AC$ as diameter. Wait, an angle inscribed in a semicircle is a right angle? Wait, no, if $AC$ is a diameter, then any angle subtended by $AC$ on the circle is a right angle. So $\angle ABC$ and $\angle ADC$ (since $B$ and $D$ are on the circle) would both be right angles? Wait, no, that can't be. Wait, no, $ABCD$ is a cyclic quadrilateral. Wait, maybe I made a mistake. Wait, the question is about the sum of $\angle ABC$ and $\angle CDA$. Let's think again.

Wait, quadrilateral $ABCD$ is cyclic (inscribed in circle $O$). So by the property of cyclic quadrilaterals, the sum of a pair of opposite angles is $180^\circ$? Wait, no, actually, the sum of all interior angles of a quadrilateral is $360^\circ$, and in a cyclic quadrilateral, opposite angles are supplementary (sum to $180^\circ$). Wait, but here, $AC$ is a diameter. So $\angle ABC$ and $\angle ADC$ (since $AC$ is diameter, $\angle ABC$ is an angle subtended by diameter $AC$, so $\angle ABC = 90^\circ$? Wait, no, an angle inscribed in a semicircle is a right angle. So if $AC$ is a diameter, then $\angle ABC$ and $\angle ADC$ (because $B$ and $D$ are on the circle) are both right angles? Wait, that would mean each is $90^\circ$, so their sum is $180^\circ$? Wait, no, that can't be. Wait, no, maybe I confused. Wait, let's re-express:

Wait, $AC$ is a diameter. So the arc $AC$ is $180^\circ$. Then $\angle ABC$ is an inscribed angle subtended by arc $AC$, so $\angle ABC = \frac{1}{2} \times 180^\circ = 90^\circ$. Similarly, $\angle ADC$ (which is $\angle CDA$) is also an inscribed angle subtended by arc $AC$, so $\angle ADC = 90^\circ$. Therefore, the sum of $\angle ABC$ and $\angle CDA$ is $90^\circ + 90^\circ = 180^\circ$? Wait, no, that can't be. Wait, no, wait, $B$ and $D$ are on opposite sides of $AC$? Wait, the diagram shows $ABCD$ as a quadrilateral with $B$ and $D$ on the circle, $AC$ as diameter. So $\angle ABC$ and $\angle ADC$ are both subtended by $AC$, but on opposite sides? Wait, no, if $AC$ is a diameter, then any point on the circle (other than $A$ and $C$) will form a right angle with $A$ and $C$. So $\angle ABC = 90^\circ$ and $\angle ADC = 90^\circ$, so their sum is $180^\circ$. Alternatively, since $ABCD$ is cyclic, the sum of $\angle ABC$ and $\angle ADC$ (since they are opposite angles? Wait, no, $ABCD$ has angles at $A$, $B$, $C$, $D$. So opposite angles are $\angle A$ and $\angle C$, $\angle B$ and $\angle D$. Wait, no, in quadrilateral $ABCD$, the opposite angles are $\angle A$ (at $A$) and $\angle C$ (at $C$), $\angle B$ (at $B$) and $\angle D$ (at $D$). Wait, but $\angle ABC$ is angle at $B$, and $\angle CDA$ is angle at $D$. So $\angle B$ and $\angle D$ are opposite angles in the cyclic quadrilateral? Wait, no, in a cyclic quadrilateral, opposite angles are supplementary. Wait, but if $AC$ is a diameter, then $\angle B$ and $\angle D$ are both right angles, so their sum is $180^\circ$. Therefore, the sum of $\angle ABC$ and $\angle CDA$ is $180^\circ$.

# Answer:
A. $180^\circ$

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