AH⊥DE This well proves that the midline and height of a regular triangle are coincident. And because EAD is perpendicular to the bottom ABCD, and AB is perpendicular to the intersection of two planes, and AB is perpendicular to the plane EAD, that is, AB⊥DE, so the plane ABH (DE⊥ Ah, DE ⊥ AB).

Let's remember point G in AB, point I in CD, and connect FG, GI and IF. Since EF is parallel and equal to AG=DI, EAD-FGI is a regular triangular prism, and F-GIBC is a quadrangular pyramid with F as the vertex and GIBC as the base. Then find the volumes of the two separately. V(ACDEF)=V(EAD-FGFI)+V(F-GIBC)

Or extend EF to point g so that FG=EF. It can be seen that ABCDEG is a regular triangular prism with EAD-GBC as the base and AB, DC and EG as the heights, and the extra volume is just F-GBC. With GBC as the base, FG is a high triangular pyramid with V(ABCDEF)=V(EAD-GBC)-V(F-GBC).