14. As shown in the figure, the area of right trapezoid $\mathrm{ABCD}$ is $52, D E=F C, \mathrm{AF}$ intersects $\mathrm{BE}$ at $\mathrm{O}$, and the area of triangle $\mathrm{AOB}$ is 17. Therefore, the sum of the areas of triangles $\mathrm{AOE}$ and $\mathrm{BOF}$ is $\qquad$.
【Answer】18
【Analysis】Algebraic analysis + area formula
Mark as shown in the right figure: Let $D E=F C=a, E F=b$, the upper base $A D=m$, the lower base $B C=n$, given $S_{\triangle A O B}=17$, let $S_{\triangle E O F}=x, S_{\triangle A O E}=y, S_{\triangle B O F}=z$,

Using the area formula for analysis:
Given: $S_{A B C D}=\frac{(m+n) \cdot(2 a+b)}{2}=52 \cdots \cdots$ (1),
As shown in the figure, $S_{\triangle A D F}=\frac{(a+b) \cdot m}{2}, S_{\triangle B E C}=\frac{(a+b) \cdot n}{2}$,
According to the problem: $S_{\triangle A D F}+S_{\triangle B E C}-x=S_{A B C D}-S_{\triangle A O B}$,
That is $\frac{(a+b) \cdot m}{2}+\frac{(a+b) \cdot n}{2}-x=52-17=35$
Organizing (1) and (2) we get: $\left\{\begin{array}{l}(m+n) \cdot(2 a+b)=104 \\ (m+n) \cdot(a+b)-2 x=70\end{array}\right.$
Subtracting (3) from (4) we get: $a \cdot(m+n)+2 x=34$, that is $\frac{a \cdot(m+n)}{2}+x=17 \cdots \cdots$ (5),
And $S_{\triangle A D E}+S_{\triangle B C F}=\frac{a m}{2}+\frac{a n}{2}=\frac{a(m+n)}{2}$,
Therefore, according to (5) we can conclude $S_{\triangle A D E}+S_{\triangle B C F}+S_{\triangle E O F}=17$,
Therefore $x+y=S_{A B C D}-S_{\triangle A O B}-\left(S_{\triangle A D E}+S_{\triangle B C F}+S_{\triangle E O F}\right)=52-17-17=18$