Area

We know that parallel lines divide lines that cut them proportionally. That is to say, let's draw another line through $A$ parallels to $DG || CF$ , then $AF = FG$ because $AP = DP$ . Let $CF$ cuts $AE$ at $Q$ , then $AQ=HQ$ .

Similarly, draw another line through $E$ parallels to $DG || CF$ , then $EH = \frac {1}{2} HQ$ because $DE = \frac {1} {2} CD$ . Since $\triangle ADE$ and $\triangle ADH$ share the same base $AE$ and height, then their areas $\frac {A_{\triangle ADH}}{A_{\triangle ADE}} = \frac {AH}{AE} = \frac {4}{5}$ .

Therefore,

$\frac {A_{\triangle ADH}}{A_{\triangle ABC}} = \frac {A_{\triangle ADH}}{A_{\triangle ADE}} \times \frac {A_{\triangle ADE}}{A_{\triangle ABC}} = \frac {AH}{AE} \times \frac {DE}{BC} = \frac {4}{5} \times \frac {1}{4} = \frac {1}{5} = \boxed {0.2}$ .