Let $K'$ be a point on the ray $AB$, such that $B$ is the midpoint of the line segment $AK'$ (Fig. 10). Then $CB$ is the median of the triangle $ACK'$. As $P$ divides this line segment in the ratio $2:1$, $P$ must be the centroid of the triangle $ACK'$. So, $K'M$, which is also a median of the triangle $ACK'$, must pass through the point $P$. Therefore, $K = K'$. So, $B$ is the midpoint of the line segment $AK$. As $AP$ passes through the point $P$, $AP$ is also a median of the triangle $ACK$. By the premises it is also an angle bisector. So, the triangle $ACK$ is isosceles with $|AC| = |AK|$ and $AP$ is its height. Therefore, $AP \perp CK$.

Fig. 10

As in solution 1, we show that $B$ is the midpoint of $AK$. So $BM$ joins the midpoints of sides in the triangle $ACK$ and therefore $BM \parallel CK$. And the angle bisector divides the opposite side in the same ratio as the corresponding sides, so $\frac{|AC|}{|AB|} = \frac{|PC|}{|BP|} = 2$. As $M$ is the midpoint of $AC$, we have $|AB| = |AM|$. An angle bisector, drawn from the vertex opposite the base in an isosceles triangle, is also the height in that triangle, giving $AP \perp BM$. It follows that $AP \perp CK$.