Problem 5. In triangle $ABC$, the lengths of the sides are three consecutive natural numbers. Let the median drawn from vertex $A$ be perpendicular to the angle bisector of $\angle ABC$. Determine the lengths of the sides of triangle $ABC$.
Solution. Triangle $ABD$ is isosceles ($BE$ is the altitude dropped from vertex $B$ and the angle bisector of $\measuredangle ABD$). It follows that $\overline{AB}=\overline{BD}=\frac{\overline{BC}}{2}$, hence $\overline{BC}=2 \overline{AB}$. There are two possibilities:

$$
\begin{aligned}
& 1^{0} \overline{BC}=\overline{AB}+2=2 \overline{AB} \text { and } \\
& 2^{0} \overline{BC}=\overline{AB}+1=2 \overline{AB}
\end{aligned}
$$

If $\overline{BC}=\overline{AB}+1=2 \overline{AB}$, then $\overline{AB}=1$ and $\overline{BC}=2$. Then $\overline{AC}=3$ or $\overline{AC}=0$. Clearly, the case where $\overline{AC}=0$ is not possible. Therefore, $\overline{AC}=3$. However, then from the triangle inequality we get $3=AB+BC>AC=3$, which is not possible. If $\overline{BC}=\overline{AB}+2=2 \overline{AB}$, then $\overline{AB}=2, \overline{BC}=4$ and $\overline{AC}=3$.

![](https://cdn.mathpix.com/cropped/2024_06_05_0c08f972e294b2535969g-09.jpg?height=544&width=339&top_left_y=1617&top_left_x=1128)

## VIII Division - 8th grade