On the sides $A B$ and $A C$ of the isosceles triangle $A B C (A B = A C)$, points $M$ and $N$ are marked respectively such that $A N > A M$. The lines $M N$ and $B C$ intersect at point $K$. Compare the lengths of segments $M K$ and $M B$.

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Since angle $ACB$ is an exterior angle for triangle $CNK$, then $\angle B = \angle C > \angle NKC$. In triangle $BKM$, the side opposite the larger angle is longer, hence, $MK > MB$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7c6a12be356bd13b62cag-05.jpg?height=666&width=695&top_left_y=21&top_left_x=676)

## Answer

$MK > MB$.