2. (10 points) In the figure, the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at $O$, and $E$ is the midpoint of $BC$. The area of triangle $ABO$ is 45, the area of triangle $ADO$ is 18, and the area of triangle $CDO$ is 69. Then the area of triangle $AED$ is . $\qquad$
【Analysis】If $A D$ is taken as the base, since point $E$ is the midpoint of $B C$, the heights of $\triangle A D B, \triangle A D E, \triangle A D C$ form an arithmetic sequence (the height of the middle triangle can be considered the average of the heights of the two side triangles), so the areas also form an arithmetic sequence (the area of the middle triangle can be considered the average of the areas of the two side triangles). This can be used to solve the problem.

【Solution】Solution: If $A D$ is taken as the base, since point $E$ is the midpoint of $B C$,
the height of $\triangle A D E$ is the average of the heights of $\triangle A D B$ and $\triangle A D C$,
therefore, the area of $\triangle A D E$ is equal to the average of the areas of $\triangle A D B$ and $\triangle A D C$.
So, $S \triangle A D E=\left(S_{\triangle A D B}+S_{\triangle A D C}\right) \div 2$
$$
\begin{array}{l}
=(45+18+18+69) \div 2 \\
=75 ;
\end{array}
$$

Answer: The area of triangle $A E D$ is 75.