This is also in MOP homework 2005--I already posted a solution in the other forum, so here goes: Let $X = AC \cap BD$, $AX = 2p$, and $BX = 2q$. Note that $AB\cdot BC = 2\cdot AD\cdot DC$ implies that $area(\triangle ABC) = 2\cdot area(\triangle ADC)$. Now, let $B', D'$ be the feet of the perpendiculars from $B,D$ to $AC$ respectively. We have $BB' = 2DD'$, and since $\triangle BXB' \sim \triangle DXD'$, we conclude that $BX = 2DX$. By power of a point, $AX\cdot XC = BX\cdot XD$, so $XC = \frac{q^2}{p}$. Because $BD = BX + XD = 3q$ and $AC = AX + XC = 2p + \frac{q^2}{p}$, AM-GM implies that \[9AC^2 = 9 \left(2p + \frac{q^2}{p}\right)^2 \geq 9(2q\sqrt{2})^2 = 72q^2 = 8BD^2,\] as desired.