In trapezoid $A B C D$, segments $A B$ and $C D$ are the bases. The diagonals of the trapezoid intersect at point $K$. Find the area of triangle $A K D$, if $A B=27, D C=18, A D=3, B C=6 \sqrt{2}$.

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Through vertex $C$, draw a line parallel to $A D$. Find the height of the resulting triangle. Determine what fraction of the area of triangle $A D B$ is the area of triangle $A K D$.

## Solution

Draw a line through point $C$, parallel to $A D$, until it intersects the base $A B$ at point $P$. Then, in triangle $P C B$, $C P = A D = 3$, $B C = 6 \sqrt{2}$,

$B P = A B - D C = 9$.

This triangle is right-angled because $B P^{2} = C P^{2} + B C^{2}$. Its height $C Q$ is found from the equality $P B \cdot C Q = C P \cdot B C$, $C Q = 2 \sqrt{2}$.

The height of triangle $A D B$ is also $2 \sqrt{2}$. Therefore, $S_{A D B} = 27 \sqrt{2}$.

From the similarity of triangles $D K C$ and $B K A$, it follows that $D K: K B = 2: 3$.

Therefore, $D K: D B = 2: 5$, $S_{A K D} = \frac{2}{5} S_{A D B} = \frac{54 \sqrt{2}}{5}$.

## Answer

$\frac{54 \sqrt{2}}{5}$.