Problem 10.8. Given a triangle $A B C$. Let point $I$ be the center of its inscribed circle, and points $P$ and $Q$ be the midpoints of sides $A B$ and $A C$ respectively. It turns out that $\angle P I Q + \angle B I C = 180^{\circ}$. Find the length of segment $B C$, if $A B = 20$ and $A C = 14$.

![](https://cdn.mathpix.com/cropped/2024_05_06_baaf26a33e2d9a36d9b4g-06.jpg?height=501&width=422&top_left_y=575&top_left_x=516)
Answer: $34 / 3$.

Solution. From the condition, it follows that $\angle B I P + \angle C I Q = 180^{\circ}$. Additionally, note that $P Q \| B C$ as the midline of the triangle.

![](https://cdn.mathpix.com/cropped/2024_05_06_baaf26a33e2d9a36d9b4g-06.jpg?height=505&width=426&top_left_y=1304&top_left_x=510)

Fig. 8: to the solution of problem 10.8

Draw a tangent to the incircle of triangle $A B C$ parallel to segment $B C$. Denote by $P^{\prime}$ and $Q^{\prime}$ the points of intersection of this tangent with sides $A B$ and $A C$ respectively (Fig. 8). Since a circle with center $I$ is inscribed in trapezoid $P^{\prime} Q^{\prime} C B$, point $I$ is the intersection of the angle bisectors of all four angles of this trapezoid. Since $P^{\prime} Q^{\prime} \| B C$, then $\angle P^{\prime} Q^{\prime} C + \angle B C Q^{\prime} = 180^{\circ}$. Therefore, $\angle I Q^{\prime} C + \angle I C Q^{\prime} = 90^{\circ}$, from which $\angle C I Q^{\prime} = 90^{\circ}$. Similarly, $\angle B I P^{\prime} = 90^{\circ}$. Then

$$
\angle B I P^{\prime} + \angle C I Q^{\prime} = 180^{\circ} = \angle B I P + \angle C I Q
$$

It is clear that if segment $P^{\prime} Q^{\prime}$ is above segment $P Q$, then $\angle B I P^{\prime} > \angle B I P$ and $\angle C I Q^{\prime} > \angle C I Q$, and if below, then $\angle B I P^{\prime} < \angle B I P$ and $\angle C I Q^{\prime} < \angle C I Q$. In both of these cases, the obtained equality is impossible. Therefore, points $P^{\prime}$ and $Q^{\prime}$ coincide with points $P$ and $Q$, i.e., quadrilateral $P Q C B$ is an inscribed trapezoid.

Note that $B P = A P = 10$ and $C Q = A Q = 7$, and also $P Q = \frac{1}{2} B C$. Thus, by the property of an inscribed quadrilateral,

$$
10 + 7 = B P + C Q = P Q + B C = \frac{3}{2} B C
$$

from which we obtain $B C = \frac{34}{3}$.

## 10th grade