Let $A B C$ be a triangle such that $\frac{|B C|}{|A B|-|B C|}=\frac{|A B|+|B C|}{|A C|}$. Determine the ratio $\angle A: \angle C$.
Answer: $1: 2$.

Denote $|B C|=a,|A C|=b,|A B|=c$. The condition $\frac{a}{c-a}=\frac{c+a}{b}$ implies $c^{2}=a^{2}+a b$ and

$$
\frac{c}{a+b}=\frac{a}{c} .
$$

Let $D$ be a point on $A B$ such that $|B D|=\frac{a}{a+b} \cdot c$ (see Figure 5). Then

$$
\frac{|B D|}{|B C|}=\frac{c}{a+b}=\frac{a}{c}=\frac{|B C|}{|B A|}
$$

so triangles $B C D$ and $B A C$ are similar, implying $\angle B C D=\angle B A C$. Also, $\frac{|A C|}{|B C|}=\frac{|A D|}{|B D|}$ yields $\frac{|B C|}{|B D|}=\frac{|A C|}{|A D|}$, and hence by the bisector theorem $C D$ is the bisector of $\angle B C A$. So the ratio asked for is $1: 2$.

![](https://cdn.mathpix.com/cropped/2024_04_17_8c3840558e5986866c27g-06.jpg?height=299&width=516&top_left_y=946&top_left_x=305)

Figure 5