Let $AC=1,\alpha=\angle CAD,\beta=\angle BAC$. Then by the law of sines, \begin{align*}AB&=\sqrt 2\sin\left(45^\circ-\beta\right),\\BC&=\sqrt 2\sin\left(\beta\right),\\CD&=\sqrt 2\sin\left(\alpha\right),\\DA&=\sqrt 2\sin\left(45^\circ-\alpha\right).\end{align*}Applying the law of cosines in $\triangle BCD$, we obtain \begin{align*}BD^2&=2\sin^2(\alpha)+2\sin^2(\beta)-4\sin(\alpha)\sin(\beta)\cos\left(90^\circ-\alpha-\beta\right)\\&=2\sin^2(\alpha)+2\sin^2(\beta)-4\sin(\alpha)\sin(\beta)\sin(\alpha+\beta).\end{align*}Using $\sin\left(45^\circ-x\right)=\frac{\sqrt 2}2\left(\cos(x)-\sin(x)\right)$ and $\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$, the condition of the problem statement becomes \begin{align*}2\sin^2(\alpha)+2\sin^2(\beta)-4\sin(\alpha)\sin(\beta)\sin(\alpha+\beta)&=BD^2=2\cdot AB\cdot BC\cdot CD\cdot DA\\&=8\sin(\alpha)\sin(\beta)\sin\left(45^\circ-\alpha\right)\sin\left(45^\circ-\beta\right)\\&=4\sin(\alpha)\sin(\beta)\left(\cos(\alpha)-\sin(\alpha)\right)\left(\cos(\beta)-\sin(\beta)\right)\\\iff \sin^2(\alpha)+\sin^2(\beta)&=2\sin(\alpha)\sin(\beta)\left[\left(\cos(\alpha)-\sin(\alpha)\right)\left(\cos(\beta)-\sin(\beta)\right)+\sin(\alpha+\beta)\right]\\&=2\sin(\alpha)\sin(\beta)\left[\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)\right]\\&=2\sin(\alpha)\sin(\beta)\cos(\alpha-\beta).\end{align*} However, by AM-GM, $2\sin(\alpha)\sin(\beta)\cos(\alpha-\beta)=\sin^2(\alpha)+\sin^2(\beta)\geq 2\sin(\alpha)\sin(\beta)$, so $\cos(\alpha-\beta)\geq 1\implies\alpha=\beta$. Thus, $D$ is the reflection of $B$ across line $AC$, which implies $AC\perp BD$, as desired.