On side $A B$ of triangle $A B C$, a point $K$ is marked. Segment $C K$ intersects the median $A M$ of the triangle at point $P$. It turns out that $A K = A P$.

Find the ratio $B K: P M$.
The first method. Draw a line through point $M$ parallel to $C K$, which intersects $A B$ at point $D$ (left figure). By Thales' theorem, $B D=K D$. By the theorem of proportional segments, $P M=K D=1 / 2 B K$.
![](https://cdn.mathpix.com/cropped/2024_05_06_c271881b12104a3b803eg-36.jpg?height=506&width=1542&top_left_y=530&top_left_x=266)

The second method. Let $T$ be the midpoint of segment $C K$ (right figure). $M T$ is the midline of triangle $C B K$, hence $M T \| B K$ and $B K=2 M T$. Triangles $K A P$ and $T M P$ are obviously similar, so $M P=M T=1 / 2 B K$.

## Answer

$B K: P M=2$.