jmc

Let $X$ and $Y$ be the points where the folded portion of the triangle crosses $AB,$ and $Z$ be the location of the original vertex $C$ after folding.



We are told that the area of $△XYZ$ is $16\%$ that of the area of $△ABC.$

Now $△ACB$ is similar to $△XZY,$ since $\angle XZY$ is the folded over version of $\angle ACB$ and since $$\angle XYZ=\angle EYB=\angle DEY=\angle CED=\angle CBA$$by parallel lines and folds. Since $△XZY$ is similar to $△ACB$ and its area is $0.16=(0.4{)}^2$ that of $△ACB,$ the sides of $△XZY$ are $0.4$ times as long as the sides of $△ACB.$

Draw the altitude of $△ACB$ from $C$ down to $P$ on $AB$ (crossing $DE$ at $Q$) and extend it through to $Z.$



Now $CP=CQ+QP=ZQ+QP=ZP+2PQ.$

Since the sides of $△XZY$ are $0.4$ times as long as the sides of $△ACB,$ then $ZP=0.4CP.$

Since $CP=ZP+2PQ,$ we have $PQ=0.3CP,$ and so $CQ=CP-PQ=0.7CP.$

Since $CQ$ is $0.7$ times the length of $CP,$ then $DE$ is $0.7$ times the length of $AB,$ again by similar triangles, so $DE=0.7(12)=\boxed{8.4}\text{ cm}.$