60. Given $\triangle A B C$. On the ray $B A$, take a point $D$ such that $|B D|=|B A|+|A C|$. Let $K$ and $M$ be two points on the rays $B A$ and $B C$ respectively such that the area of $\triangle B D M$ is equal to the area of $\triangle B C K$. Find $\widehat{B K M}$, if $\widehat{B A C}=\alpha$.
60. The condition $S_{\triangle B D M}=S_{\triangle B C K}$ means that

or

$$
|B D| \cdot |B M| = |B K| \cdot |B C|
$$

$$
(|B A| + |A C|) |B M| = |B K| \cdot |B C| .
$$

Draw a line through $M$ parallel to $A C$; let $L$ be the point of intersection of this line with $B A$. We will prove that $|L M| = |K L|$; from this it will follow that the desired angle $\widehat{B K M} = \frac{1}{2} \cdot \widehat{B A C} = \frac{\alpha}{2}$. Since $\triangle B L M$ and $\triangle B A C$ are similar, then

$$
|L M| = \frac{|B M|}{|B C|} \cdot |A C|, \quad |B L| = \frac{|B M|}{|B C|} \cdot |A B| .
$$

Now find $|B K|$ from (1) and calculate $|K L| = |B K| - |B L|$: $=$ $=\frac{|B A| + |A C|}{|B C|} \cdot |B M| - \frac{|B M|}{|B C|} \cdot |A B| = \frac{|B M|}{|B C|} \cdot |A C|$, i.e., indeed $|L M| = |K L|$.

Answer: $\alpha / 2$.