A straight line parallel to the base BC of the triangle ABC intersects $\overline{\mathrm{AB}}\text{ and }\overline{\mathrm{AC}}$ at the points B and E respectively. If the area of the triangle ACD is  $36{\mathrm{cm}}^2$, then the area of  $\mathrm{△}\mathrm{ACD}$

DE||BC 
$\mathrm{△}DEB\text{ and }\mathrm{△}DEC are the triangle with the same base and between same parallel line$
 

$$\begin{gathered} \therefore \mathrm{Ar}\left(\mathrm{\Delta DEB}\right)=\mathrm{Ar}\left(\mathrm{\Delta DEC}\right) \left(1\right) \\ \mathrm{Given} \mathrm{Ar}\left(\mathrm{△}\mathrm{ABE}\right)=36{\mathrm{cm}}^2 \\ \Rightarrow \mathrm{Ar}\left(\mathrm{△}\mathrm{ADE}\right)+\mathrm{Ar}\left(\mathrm{\Delta DEB}\right)=36{\mathrm{cm}}^2(\therefore \text{ from }(1\left)\right) \\ \Rightarrow \mathrm{Ar}\left(\mathrm{\Delta ACD}\right)=36{\mathrm{cm}}^2 \end{gathered}$$