Thales' Theorem and the Proportional Segments Theorem [The ratio of the areas of triangles with a common base or common height]

In triangle $ABC$, a point $K$ is taken on side $AB$, such that $AK: BK=1: 2$, and a point $L$ is taken on side $BC$, such that $CL: BL=2: 1$. $Q$ is the point of intersection of lines $AL$ and $CK$. Find the area of triangle $ABC$, given that $S_{BQC}=1$.
Draw a line through point $L$ parallel to $C K$, until it intersects side $A B$ at point $M$. By Thales' theorem, $K M=2 / 3 B K=4 / 3 A K=4 / 7 A M$. Therefore, $S_{A B C}={ }^{A L} / Q L \cdot S_{B Q C}={ }^{A M} / K M \cdot 1=7 / 4$.

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