4. In triangle $\mathrm{ABC}$ with sides $a, b$ and $c$, it is known that $b=50 \mathrm{~cm}, c=48 \mathrm{~cm}$ and the internal angle at vertex A is twice the internal angle at vertex B. Determine the side $a$.
Solution: Extend side CA through vertex A to point D such that $\overline{\mathrm{AD}}=\overline{\mathrm{AB}}$. Then, triangle DBA is isosceles, so $\triangle D=\triangle D B A$. For this triangle, $\triangle B A C=2 \beta$ is an exterior angle. It is equal to the sum of the base angles

![](https://cdn.mathpix.com/cropped/2024_06_05_e7266f2e82216ac5d245g-5.jpg?height=644&width=411&top_left_y=252&top_left_x=1049)
$D B$, i.e., $\triangle D+\triangle D B A=2 \beta$. Therefore, $\triangle D=\triangle D B A=\beta$. However, then $\triangle C B D=2 \beta$, so triangles $\mathrm{ABC}$ and BDC are similar. We have $\overline{\mathrm{BC}}: \overline{\mathrm{AC}}=\overline{\mathrm{CD}}: \overline{\mathrm{BC}}$ or $a: 50=(50+48): a ; a^{2}=50 \cdot 48$; i.e., $a=70 \mathrm{~cm}$.