2. In triangle $ABC$, we have $\overline{AB}=2 \text{~cm}, \overline{BC}=4 \text{~cm}$, and $\overline{CA}=2 \sqrt{2} \text{~cm}$. Let $P$ be a point on the angle bisector of angle at vertex $B$ such that $AP$ is perpendicular to that bisector, and let $Q$ be a point on the angle bisector of angle at vertex $C$ such that $AQ$ is perpendicular to that bisector. Calculate the length of the segment $PQ$.
Solution. Let $AP$ intersect $BC$ at point $M$, and $AQ$ intersect $BC$ at point $N$. Then triangles $ABM$ and $ACN$ are isosceles, so

![](https://cdn.mathpix.com/cropped/2024_06_05_2d390a2087465305abc8g-3.jpg?height=266&width=553&top_left_y=1977&top_left_x=913)
therefore $\overline{BM}=2 \mathrm{~cm}$ and $\overline{CN}=2 \sqrt{2} \mathrm{~cm}$. From

$$
\overline{BM}+\overline{CN}-\overline{NM}=\overline{BC}
$$

It follows that,

$$
\overline{NM}=\overline{BM}+\overline{CN}-\overline{BC}=2+2 \sqrt{2}-4=2 \sqrt{2}-2 \mathrm{~cm}
$$

Since $PQ$ is the midline of triangle $ANM$, it follows that

$$
\overline{PQ}=\frac{2 \sqrt{2}-2}{2}=\sqrt{2}-1 \mathrm{~cm}
$$