In a cyclic quadrilateral $ABCD$ inscribed in a circle, $AB \| CD$. Find a relationship between the angles $\angle CBA$ and $\angle CAB$.

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Let the angle $C A B$ be denoted by $\alpha$, the angle $C B A$ by $\beta$, and let $A B = a$ and $C D = b$.

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Since the trapezoid is a cyclic trapezoid, the opposite angles satisfy $A B C \varangle + C D A \varangle = 180^{\circ}$. However, the sum of $A B C \varangle + B C D \varangle$ is also $180^{\circ}$, and therefore $C D A \varangle = B C D \varangle$, which means the trapezoid is isosceles: $D A = C B = c$. The perpendicular from $C$ to the base $A B$ intersects $A B$ at point $T$, which is either a point on the segment $A B$, in which case

$$
B T = \frac{a - b}{2} \quad \text{and} \quad A T = a - \frac{a - b}{2} = \frac{a + b}{2}
$$

or it falls on the extension of segment $A B$ beyond $B$, in which case

$$
B T = \frac{b - a}{2} \quad \text{and} \quad A T = a + \frac{b - a}{2} = \frac{a + b}{2}
$$

From the right triangles $A C T$ and $B C T$,

$$
\operatorname{tg} \alpha = \frac{C T}{A T} = \frac{2 C T}{a + b}
$$

and

$$
\sin \beta = \sin \left(180^{\circ} - \beta\right) = \frac{C T}{C B} = \frac{2 C T}{2 c}
$$

Our trapezoid is also a tangential trapezoid, so the sum of the opposite sides is equal: $a + b = 2 c$, from which we can deduce the relationship between the angles:

$$
\operatorname{tg} \alpha = \sin \beta
$$

Remark. The obtained relationship is reversible in the following sense: if in the isosceles trapezoid $A B C D$, $\operatorname{tg} C A B \varangle = \sin C B A \varangle$, then the trapezoid is both a cyclic and a tangential trapezoid.