4. In triangle $ABC$, the angle at vertex $C$, i.e., $\measuredangle ACB$ is $48^{\circ}$. The normal to the angle bisector of $\measuredangle ACB$ passing through vertex $C$ intersects line $AB$ at point $D$, such that point $B$ lies between points $A$ and $D$ and $\overline{AD}=\overline{AC}+\overline{BC}$. Calculate the measures of the angles in triangle $ABC$.
Solution. The line $CD$ which is normal to the bisector of the internal $\measuredangle ACB$ is simultaneously the bisector of the external angle at vertex $C$. We extend the side $AC$ beyond vertex $C$ and mark the point $E$ such that $\overline{BC}=\overline{CE}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_b36983d4a97bcdb99a36g-04.jpg?height=259&width=511&top_left_y=1351&top_left_x=951)
The triangles $BCD$ and $CDE$ are congruent by the SAS (Side-Angle-Side) congruence criterion. From their congruence, it follows that $\measuredangle CBD = \measuredangle CED$, and $\measuredangle CBD = \alpha + \gamma$. Since

$$
\overline{AE} = \overline{AC} + \overline{CE} = \overline{AC} + \overline{BC} = \overline{AD}
$$

it follows that triangle $ADE$ is isosceles with angles

$$
\measuredangle ADE = \measuredangle AED = \alpha + \gamma.
$$

For triangle $ADE$, we have $\alpha + 2(\alpha + \gamma) = 180^{\circ}$, and since $\gamma = 48^{\circ}$, we get $\alpha = 28^{\circ}$ and $\beta = 104^{\circ}$.

## VIII Section