Problem 5. Given a triangle $ABC$ with $\measuredangle CBA=20^{\circ}, \measuredangle ACB=40^{\circ}$ and $\overline{AD}=2 \text{~cm}$, where $D$ is the intersection point of the angle bisector of the angle at vertex $A$ and side $BC$. Determine the difference $\overline{BC}-\overline{AB}$.
Solution. On side $BC$, we choose a point $E$ such that $\measuredangle AEB=80^{\circ}$. Then $\measuredangle BAE=80^{\circ}$, so triangle $ABE$ is isosceles, i.e., $\overline{AB}=\overline{AE}$. Triangle $AEC$ is isosceles because $\measuredangle EAC=40^{\circ}$, so $\overline{AE}=\overline{CE}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_1cb7beb611b796adc950g-08.jpg?height=285&width=927&top_left_y=653&top_left_x=457)

Triangle $ADE$ is isosceles because $\measuredangle ADE=80^{\circ}$, so $\overline{AE}=\overline{AD}=$ $2 \text{ cm}$. Therefore, $\overline{BC}-\overline{AB}=\overline{BC}-\overline{BE}=\overline{CE}=\overline{AE}=\overline{AD}=2 \text{ cm}$

## VII Section