In triangle $A B C$, a point $K$ is taken on side $A C$, such that $A K=1, K C=3$, and a point $L$ is taken on side $A B$, such that $A L: L B=2: 3$. Let $Q$ be the intersection point of lines $B K$ and $C L$. The area of triangle $A Q C$ is 1. Find the height of triangle $A B C$, dropped from vertex $B$.
Draw a line through point $K$ parallel to $CL$, until it intersects side $AB$ at point $M$. By Thales' theorem, $AM: ML = AK: KC = 1: 3$. Therefore,

$ML = 3/4 AL = 1/2 LB$. Again, by Thales' theorem, $KQ = 1/2 BQ = 1/3 BK$.

Thus, $S_{ABC} = 3 S_{AQC} = 3$, and the desired height is ${ }^{2 S_{ABC} / AC} = 3 / 2$.

## Answer

1.5.