Problem 8.8. Given a triangle $A B C$, in which $\angle A=42^{\circ}$ and $A B<A C$. Point $K$ on side $A C$ is such that $A B=C K$. Points $P$ and $Q$ are the midpoints of segments $A K$ and $B C$ respectively. How many degrees does angle $A C B$ measure, if it is known that $\angle P Q C=110^{\circ} ?$

![](https://cdn.mathpix.com/cropped/2024_05_06_191a5994d6b917032c66g-06.jpg?height=339&width=551&top_left_y=95&top_left_x=451)
Answer: 49.

![](https://cdn.mathpix.com/cropped/2024_05_06_191a5994d6b917032c66g-06.jpg?height=295&width=797&top_left_y=551&top_left_x=332)

Fig. 4: to the solution of problem 8.8

Solution. Mark a point $L$ on the extension of side $C A$ beyond point $A$ such that $A L = A B$ (Fig. 4).

Notice that line $P Q$ is parallel to line $B L$ as the midline in triangle $B C L$. Therefore, $\angle L B C = \angle P Q C = 110^{\circ}$.

In the isosceles triangle $B A L$, the external angle at vertex $A$ is $42^{\circ}$, so the angles at the base are $\angle A L B = \angle A B L = \frac{1}{2} \cdot 42^{\circ} = 21^{\circ}$.

From the sum of the angles in triangle $B C L$, we find the required angle:

$$
\angle A C B = 180^{\circ} - \angle C L B - \angle L B C = 180^{\circ} - 110^{\circ} - 21^{\circ} = 49^{\circ}
$$

## 8th grade