Question 13
In a triangle $ABC$, $D$ is a point on $BC$ and $F$ is a point on $AB$. The point $K$ of reflection of $B$ across $DF$ is on the opposite side of $AC$ to $B$. $AC$ intersects $FK$ at $P$ and $DK$ at $Q$. The total area of triangles $AFP$, $PKQ$ and $QDC$ is $10\,\mathrm{cm}^2$. If we add to this the area of the quadrilateral $DFPQ$, we obtain $\frac{2}{3}$ of the area of $ABC$. What is the area, in $\mathrm{cm}^2$, of triangle $ABC$?
