Problem 5. On the lateral sides $AB$ and $BC$ of an isosceles triangle $ABC$, points $M$ and $N$ are marked such that $AM = MN = NC$. On the side $AC$, points $P$ and $Q$ are chosen such that $MQ \parallel BC$ and $NP \parallel AB$. It is known that $PQ = BM$. Find the angle $MQB$.
Answer: $36^{\circ}$. The answer can also be given in the form $\arccos \left(\frac{\sqrt{5}+1}{4}\right)$, etc.

Solution. Let $\angle A=\angle C=\alpha$, then by the property of parallel lines $\angle N P C=$ $\angle A Q M=\alpha$. It is not difficult to prove that $M N \| A C$ (through the similarity of triangles, or through the equality of distances from points $M, N$ to the line $A C$, or through the calculation of angles), therefore $\angle B M N=$ $\angle Q M N=\angle B N M=\angle M N P=\alpha$ and $A M N P$ is a rhombus, from which $A P=A M=M N$. (Seventh-graders can prove this equality through the equality $\triangle A M N=\Delta A P N$.)

![](https://cdn.mathpix.com/cropped/2024_05_06_4b75cccb2916eb1c9e18g-2.jpg?height=368&width=759&top_left_y=684&top_left_x=206)

First case. Suppose point $P$ lies on segment $A Q$. Then, taking into account the above, $\triangle A M Q=\triangle A P B$ by two sides and the angle between them, so $\angle A B P=\angle A Q M=\alpha$; similarly, $\angle Q B C=\angle C P N=\alpha$. $\angle B P Q=2 \alpha$ as an external angle for $\triangle A P B$, hence $\angle B P N=2 \alpha-\alpha=$ $\alpha$; similarly, $\angle M Q B=\alpha$. Since $\triangle A B Q$ is isosceles, $\angle A B Q=\angle A Q B=2 \alpha$, from which $\angle P B Q=\alpha$. By the theorem on the sum of angles in a triangle for $\triangle A B C$ we get $5 \alpha=180^{\circ}$, hence $\angle M Q B=\alpha=36^{\circ}$.

Second case. Suppose point $P$ lies on segment $C Q$. Then $M N=A P>P Q=B M$. In $\triangle B M N$ the larger angle lies opposite the larger side, so $\angle M B N=180^{\circ}-2 \alpha>\angle B N M=\alpha$, from which $\alphaA Q=A P-Q P=A M-Q P$, this means that $\angle A=\alpha>\angle A M Q=180^{\circ}-2 \alpha$, from which $\alpha>60^{\circ}$. The obtained contradiction shows that the second case is impossible.