3. In triangle $ABC$, the angles at vertices $B$ and $C$ are equal to $40^{\circ}$. Let $D$ be a point on line $AB$ such that point $B$ is between points $A$ and $D$ and $AD = BC$. Determine the angles of triangle $ADC$.
Solution. Let $E$ be a point on the side of line $BC$ where point $A$ is not, such that triangle $BCE$ is equilateral, as shown in the diagram on the right. Since $\measuredangle CAD = \measuredangle ECA = 100^{\circ}$ and $AD = BC = EC$, we conclude that quadrilateral $DECA$ is an isosceles trapezoid. Therefore, $\measuredangle ECD = 80^{\circ} = \measuredangle EBD$, so $EB = ED$. Thus, point $E$ is the center of the circle circumscribed around triangle $BCD$. From this, it follows

![](https://cdn.mathpix.com/cropped/2024_06_05_b1b297325d2ee7aabbcfg-1.jpg?height=533&width=524&top_left_y=1668&top_left_x=940)

$$
\measuredangle BCD = \frac{1}{2} \measuredangle BED = \frac{1}{2} \cdot 20^{\circ} = 10^{\circ} \text{.}
$$

Accordingly, the angles of triangle $ADC$ are $100^{\circ}$, $30^{\circ}$, and $50^{\circ}$.