12. As shown in the figure, in quadrilateral $\mathrm{ABCD}$, $\mathrm{AC}$ and $\mathrm{BD}$ intersect at point $\mathrm{O}$. Draw the altitude $\mathrm{DE}$ of triangle $\mathrm{DBC}$, and connect $\mathrm{AE}$. If the area of triangle $\mathrm{ABO}$ is equal to the area of triangle $\mathrm{DCO}$, and $\mathrm{DC}=17$ cm, $\mathrm{DE}=15$ cm, then what is the area of the shaded part in square centimeters?
【Analysis】Since $S_{\triangle A B O}=S_{\triangle D C O}$, then $S_{\triangle A B C}=S_{\triangle D C B}$. Because the two triangles share the base $\mathrm{BC}$, the heights from $\mathrm{BC}$ in both triangles are equal, so $\mathrm{AD}$ is parallel to $\mathrm{BC}$. Therefore, in triangle $\mathrm{ACE}$, the height on side $\mathrm{CE}$ is 15 cm.
In the right triangle $\mathrm{CDE}$, by the Pythagorean theorem, we have
$$
C E^{2}=C D^{2}-D E^{2}=17^{2}-15^{2}=(17+15)(17-15)=64 \text {, }
$$

Thus, $\mathrm{CE}=8$ cm.
Therefore, $S_{\triangle A C E}=\frac{1}{2} \times 8 \times 15=60$ square cm.